How many three-digit even numbers can be formed using the digits 1, 2, 3, 4 and 5, when repetition of digits is not allowed?
A36
B30
C24
D12
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The Setup: We need to build a 3-digit number from the set {1,2,3,4,5} without any repeating digits. To be an even number, the last digit has to pass the vibe check (it must be divisible by 2).
Step 1: Fix the units digit. The only even digits in our squad are 2 and 4. That gives us 2 valid options for the final slot.
Step 2: Fill the remaining slots. We have a 3-digit number (Hundreds, Tens, Units).
We used 1 digit for the units place, leaving 4 digits available.
For the Hundreds place, we have 4 options.
For the Tens place, we have 3 options.
Step 3: Multiply them together using the Fundamental Principle of Counting.
Total=4×3×2=24Final Answer: 24
The number of triangles that can be formed by choosing points from 7 points on a line and 5 points on another parallel line is _________.
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The Setup: We are choosing points to form triangles from two parallel lines containing 7 and 5 points respectively.
Step 1: Identify the combinatorial cases for triangle formation.
A triangle requires exactly 3 non-collinear vertices. Since the points lie on two parallel lines, taking 3 points from the same line is invalid. Thus, we must select either:
* Case 1: 1 point from the first line and 2 points from the second line.
* Case 2: 2 points from the first line and 1 point from the second line.
Step 2: Calculate configurations for Case 1.
Choose 1 point from the line with 7, and 2 points from the line with 5:
Ways1=(17)×(25)=7×(25×4)=7×10=70Step 3: Calculate configurations for Case 2.
Choose 2 points from the line with 7, and 1 point from the line with 5:
Ways2=(27)×(15)=(27×6)×5=21×5=105Step 4: Sum the valid triangle configurations.
Total Triangles=70+105=175Final Answer: 175
There are 5 parallel lines on the plane. On the same plane, there are n other lines that are perpendicular to the 5 parallel lines. If the number of distinct rectangles formed by these lines is 360, what is the value of n?
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The Setup: We are stepping into the Combinatorics arena. There are 5 parallel lines and n perpendicular lines. To build a rectangle, you need a squad of exactly 4 lines: 2 horizontal and 2 vertical. It's like picking your duo for Fortnite-you need exactly two from each distinct set to lock in the shape.
Step 1: Calculate the number of pairs we can pull from the 5 parallel lines. Choosing 2 out of 5 is basic combinations:
(25)=25×4=10Step 2: Set up the equation for the perpendicular lines. We need 2 lines from the n squad, which gives us (2n). The total number of distinct rectangles is 360.
10×(2n)=360Step 3: Solve for n. Divide both sides by 10 to strip it down:
(2n)=362n(n−1)=36n(n−1)=72Step 4: Find the consecutive integers that multiply to 72. Since 9×8=72, we know n=9. No cap, it's that clean.
Final Answer: 9
In a chess tournament, there are four groups, each containing an equal number of players. Each player plays
1. against every other player belonging to one's own group exactly once;
2. against each player belonging to one of the remaining three groups exactly twice;
3. against each player belonging to one of the remaining two groups exactly three times; and
4. against each player belonging to the remaining group exactly four times.
If there are more than 1000 matches being played in the tournament, the minimum possible number of players in each group is
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The Setup: A chess tournament consists of four equal groups. Players play matches inside their group and against the other three groups at varying frequencies (1×,2×,3×,4×). We must find the minimum group size to surpass 1000 total matches.
Step 1: Analyze the match distribution logic.
Let each of the 4 groups have k players. Total players = 4k.
Every group must be assigned an inter-group match frequency of 2,3, and 4 against the other three groups.
This is perfectly modeled by weighting the edges of a K4 graph with values {2,3,4} such that every vertex touches exactly one of each weight. Such a configuration requires three distinct perfect matchings on K4, meaning the edge weights between the 6 pairs of groups are exactly: two 2s, two 3s, and two 4s.
Step 2: Calculate total inter-group matches.
Between any two groups of size k playing each other w times, the number of matches is k×k×w=wk2.
Summing over the 6 pairs of groups:
Inter-Group Matches=k2(2+2+3+3+4+4)=18k2Step 3: Calculate total intra-group matches.
Each player plays everyone in their own group exactly once.
For one group, the matches are (2k)=2k(k−1).
Since there are 4 groups:
Intra-Group Matches=4×2k(k−1)=2k(k−1)=2k2−2kStep 4: Formulate the total matches inequality.
Total Matches=18k2+2k2−2k=20k2−2k
We are given that there are more than 1000 matches:
20k2−2k>1000⟹10k2−k>500Step 5: Test integer values for k to find the minimum.
* If k=7: 10(49)−7=490−7=483 (Fails, 483≯500)
* If k=8: 10(64)−8=640−8=632 (Succeeds, 632>500)
The minimum possible number of players in each group is 8.
Final Answer: 8
The total number of positive integer solutions of 21≤a+b+c≤25 is
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The Setup: We must find the total number of positive integer solutions to the bounded inequality 21≤a+b+c≤25.
Step 1: Normalize the variables to utilize Stars and Bars.
Since a,b,c must be strictly positive integers (≥1), let:
a=x+1b=y+1c=z+1
where x,y,z are non-negative integers (≥0).
Substitute these into the inequality:
21≤(x+1)+(y+1)+(z+1)≤2521≤x+y+z+3≤2518≤x+y+z≤22Step 2: Apply the combinatorics formula for exact sums.
The number of non-negative integer solutions to x+y+z=n is given by (3−1n+3−1)=(2n+2).
We need to sum this over n∈{18,19,20,21,22}:
Solutions=(220)+(221)+(222)+(223)+(224)Step 3: Evaluate the sum.
Calculate each combination (2k)=2k(k−1):
(220)=220×19=190(221)=221×20=210(222)=222×21=231(223)=223×22=253(224)=224×23=276Total=190+210+231+253+276=1160Final Answer: 1160
Out of 13 objects, 4 are indistinguishable and the rest are distinct. The number of ways we can choose 4 objects out of 13 objects is __________.
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The Setup: This combinatorics problem mixes indistinguishable (identical) objects with distinct ones. The standard combination formula (rn) only works for choosing distinct items. When dealing with indistinguishable objects, choosing one of them gives the same exact outcome as choosing another, so there is only 1 way to select any given quantity of them. We must break this down into cases based on how many identical items we draft into the final squad of 4.
Step 1: Catalog the inventory. Total pool =13 objects.
Indistinguishable objects =4. Distinct objects =13−4=9.
We need to form a team of exactly 4 objects.
Step 2: Calculate combinations case-by-case. We iterate through every possible quantity of indistinguishable objects, from 0 up to 4 (we cannot take more than the 4 that exist).
* Case 1 (0 identical, 4 distinct):(49)=4×3×2×19×8×7×6=126 ways
* Case 2 (1 identical, 3 distinct):1×(39)=3×2×19×8×7=84 ways
* Case 3 (2 identical, 2 distinct):1×(29)=2×19×8=36 ways
* Case 4 (3 identical, 1 distinct):1×(19)=9 ways
* Case 5 (4 identical, 0 distinct):1×(09)=1 way
Step 3: Aggregate the total possibilities. The cases are mutually exclusive (they differ in how many identical objects were taken), so we add them:
126+84+36+9+1=256Step 4: Cross-check with a symmetry argument. That total is (09)+(19)+(29)+(39)+(49) - the lower half of row 9 of Pascal's triangle. Since (k9)=(9−k9), the ten entries of that row pair off perfectly into two equal halves, so this sum is exactly half the row total:
229=2512=256
Two independent routes, same number, so the count is confirmed.
Final Answer: 256
Mrs and Mr Sharma, and Mrs and Mr Ahuja along with four other persons are to be seated at a round table for dinner. If Mrs and Mr Sharma are to be seated next to each other, and Mrs and Mr Ahuja are not to be seated next to each other, then the total number of seating arrangements is _________.
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The Setup: A combinatorics problem where 8 individuals are arranged at a round table. We apply circular permutation blocks and complementary counting to satisfy simultaneous grouping constraints (one pair together, one pair separated).
Step 1: Calculate permutations satisfying the first constraint.
Total people = 8. Constraint 1: The Sharma pair must sit together.
Bind the Sharmas into a single block. We now have 7 distinct entities to arrange in a circle (the Sharma block + 6 others).
Circular arrangements=(n−1)!=(7−1)!=6!
The Sharmas can also swap seats internally within their block (2!).
Total ways (Sharmas together)=6!×2=720×2=1440Step 2: Calculate overlapping violations (both pairs together).
Constraint 2: The Ahuja pair must NOT sit together. We find the restricted subset where they *do* sit together to subtract it from our total.
Bind the Sharmas into one block and the Ahujas into another block. We now have 6 distinct entities to arrange.
Circular arrangements=(6−1)!=5!
Both the Sharma block and the Ahuja block have internal swaps (2!×2!).
Ways (Both pairs together)=5!×2×2=120×4=480Step 3: Apply complementary counting.
Subtract the violating cases from the initial grouping constraints:
Valid Seating=(Sharmas together)−(Both pairs together)Valid Seating=1440−480=960Final Answer: 960
Let n be the number of ways in which 20 identical balloons can be distributed among 5 girls and 3 boys such that everyone gets at least one balloon and no girl gets fewer balloons than a boy does. Then
A9000≤n<10000
B8000≤n<9000
C7000≤n<8000
D6000≤n<7000
Pick an option to attempt
This question was cancelled by the exam board. No option was accepted as correct and all candidates were awarded full marks.
The Setup: There are 20 identical balloons to be distributed among 5 girls and 3 boys such that everyone receives at least one balloon. A condition specifies that no girl receives fewer balloons than any boy does.
Step 1: Translate the logical constraints mathematically.
Let Gi be the balloons received by girl i and Bj be the balloons received by boy j.
The condition implies min(Gi)≥max(Bj). Let M=max(Bj). Since Bj≥1, M≥1.
The boys must consume at least M+2 balloons (one gets M, the others at least 1).
The girls must consume at least 5M balloons.
Total bounded constraint: 5M+M+2≤20⟹6M≤18⟹M≤3.
Step 2: Analyze valid configurations partitioned by M.
**Case 1: M=1.**
All 3 boys get 1 balloon (1 way). The 5 girls distribute the remaining 17 balloons (Gi≥1).
Using stars and bars: (5−117−1)=(416)=1820.
**Case 2: M=2.**
Boy combinations with a max of 2: (2,2,2) [1 way], (2,2,1) [3 ways], (2,1,1) [3 ways].
Sum(B)=6, remaining 14 to girls (Gi≥2). Ways =1×70=70.
Sum(B)=5, remaining 15 to girls (Gi≥2). Ways =3×126=378.
Sum(B)=4, remaining 16 to girls (Gi≥2). Ways =3×210=630.
Total for Case 2 =70+378+630=1078.
**Case 3: M=3.**
Boys must sum to ≤5 since girls need at least 5×3=15. The only boy configuration is (3,1,1) [3 ways].
Sum(B)=5, remaining 15 to girls (Gi≥3). Ways =3×1=3.
Step 3: Sum the cases.
Total exact valid distributions =1820+1078+3=2901.
Note: All four options give ranges from 6000 upward, so none of them contains 2901. This question was cancelled by the exam board and all candidates were awarded full marks. The value 2901 is what the stated constraints actually yield.
Final Answer: 2901
The number of 5-digit numbers consisting of distinct digits that can be formed such that only odd digits occur at odd places is
A5250
B6240
C2520
D3360
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The Setup: Permutations and Combinations puzzle with strict placement rules. We are building a 5-digit number (Positions 1, 2, 3, 4, 5). The odd positions (1st, 3rd, 5th) are strictly reserved for odd digits.
Step 1: Draft the odd squad. We have exactly 5 odd digits available (1, 3, 5, 7, 9). We need to select 3 of them and arrange them into the 3 odd slots.
Number of ways to fill odd places = P(5,3)=5×4×3=60 ways.
Step 2: Assess the remaining free agents. We have 5 even digits (0, 2, 4, 6, 8) and 2 leftover odd digits that didn't get drafted into the odd slots. That gives us a pool of 7 distinct digits.
Step 3: Draft for the even slots (2nd and 4th). The prompt says 'only odd digits occur at odd places', which means odd places are restricted, but even places are fully unrestricted. They can accept any remaining digit, including zero, because the 1st digit is already guaranteed to be a non-zero odd digit.
We select 2 digits from the pool of 7 and arrange them.
Number of ways to fill even places = P(7,2)=7×6=42 ways.
Step 4: Multiply the independent events to find the total combinations.
Total Numbers=60×42=2520Final Answer: 2520
In how many ways can the letters of the word MANAGEMENT be arranged such that no two vowels appear together?
A75600
B25200
C37800
D21600
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The Setup: We must arrange the 10 letters of MANAGEMENT so that no two vowels are adjacent. This relies on the Gap Method of combinatorics.
Step 1: Audit the frequency of the letters.
The word MANAGEMENT consists of 10 letters:
Vowels (4 total): A (2), E (2).
Consonants (6 total): M (2), N (2), G (1), T (1).
Step 2: Arrange the consonants.
First, place the 6 consonants in a row.
The number of unique permutations accounting for identical letters is:
2!×2!6!=4720=180 waysStep 3: Place the vowels in the gaps.
Arranging 6 consonants in a line creates exactly 7 gaps (including the outside ends) where vowels can be slotted to ensure they never touch.
We must choose 4 gaps for our 4 vowels:
(47)=35 ways
Next, arrange the 4 vowels (A, A, E, E) within those 4 chosen gaps:
2!×2!4!=424=6 waysStep 4: Calculate the total configurations.
Multiply the independent combinatorics together:
Total Ways=180×35×6=180×210=37800Final Answer: 37800
There are 10 points in the plane, of which 5 points are collinear and no three among the remaining are collinear. Then the number of distinct straight lines that can be formed out of these 10 points is
A10
B25
C35
D36
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The Setup: Geometric Combinatorics. To draw a straight line, you literally just need to connect 2 points. Normally, you'd calculate all possible pairs. But when points are collinear (sitting on the exact same line), connecting any two of them just redraws the exact same line. We have to filter out the duplicates.
Step 1: Calculate the raw, unfiltered max number of lines. Choosing any 2 points out of 10 gives us:
(210)=210×9=45 possible pairsStep 2: Calculate the overlapping spam. The 5 collinear points are generating redundant lines. If we picked pairs exclusively from these 5 points, we'd get:
(25)=25×4=10 redundant pairsStep 3: Clean up the geometry. We subtract all 10 redundant pairs from our total because they don't form unique lines. However, we still have to add 1 back in to represent the single, massive straight line that all 5 of those points are sitting on.
Total Distinct Lines=45−10+1=36Final Answer: 36
If 5 boys and 3 girls sit randomly around a circular table, the probability that there will be at least one boy sitting between any two girls is
A71
B72
C53
D41
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The Setup: There are 5 boys and 3 girls sitting randomly around a circular table. We must find the probability that there is at least one boy sitting between any two girls.
Step 1: Calculate the total number of unrestricted circular arrangements.
For n distinct individuals sitting in a circle, the number of unique arrangements is (n−1)!.
Total people =5+3=8.
Total Arrangements=(8−1)!=7!=5040Step 2: Calculate the number of restricted arrangements (girls separated).
First, arrange the 5 boys in a circle.
Ways to seat boys=(5−1)!=4!=24
Seating the 5 boys creates exactly 5 gaps between them. To ensure no two girls sit adjacent to each other, we place the 3 girls into these distinct gaps.
Choose 3 gaps out of 5, and arrange the girls:
Ways to seat girls=(35)×3!=10×6=60
Total valid arrangements =24×60=1440.
Step 3: Calculate the probability.
Probability=50401440=504144=4212=72Final Answer:72
Consider an 8×8 chessboard. The number of ways 8 rooks can be placed on the board such that no two rooks are in the same row and no two are in the same column is
A7
B7!
C8
D8!
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The Setup: We need to find the total number of ways to place 8 indistinguishable rooks on a standard 8×8 chessboard such that no two rooks threaten each other (no two share the same row or column).
Step 1: Place the rooks sequentially row by row.
To ensure no two rooks share a row, exactly one rook must be placed in each of the 8 rows.
* Row 1: The first rook can be placed in any of the 8 squares (columns) in the first row.
* Row 2: The second rook must be placed in the second row, but it cannot share the column occupied by the first rook. This leaves 7 valid squares.
* Row 3: The third rook cannot share a column with the first two rooks, leaving 6 valid squares.
* …
* Row 8: The final rook is forced into the single remaining unoccupied column.
Step 2: Calculate total configurations.
The total number of valid placements is the product of the independent choices for each row:
Total Ways=8×7×6×5×4×3×2×1=8!Final Answer: 8!
The number of solutions of the equation x1+x2+x3+x4=50, where x1,x2,x3,x4 are integers with x1≥1,x2≥2,x3≥0,x4≥0 is
A20200
B19200
C19600
D18400
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The Setup: We are asked to find the number of integer solutions to a linear equation subject to specific lower-bound constraints. We will use a variable substitution technique to normalize the lower bounds to zero, allowing the application of the stars and bars combinatorial method.
Step 1: Define the equation and initial constraints.
Equation: x1+x2+x3+x4=50
Constraints: x1≥1,x2≥2,x3≥0,x4≥0Step 2: Normalize the variables to zero-bounded equivalents.
Let y1=x1−1, which ensures y1≥0.
Let y2=x2−2, which ensures y2≥0.
Let y3=x3, keeping y3≥0.
Let y4=x4, keeping y4≥0.
Step 3: Substitute the normalized variables back into the original equation.
(y1+1)+(y2+2)+y3+y4=50y1+y2+y3+y4+3=50y1+y2+y3+y4=47Step 4: Apply the stars and bars formula.
The number of non-negative integer solutions to y1+y2+⋯+yk=n is given by (k−1n+k−1).
Here, n=47 and k=4:
Solutions=(4−147+4−1)=(350)Step 5: Evaluate the binomial coefficient.
(350)=3×2×150×49×48=50×49×8=19600Final Answer: 19600
A rabbit is sitting at the base of a staircase which has 10 steps. It proceeds to the top of the staircase by climbing either one step at a time or two steps at a time. The number of ways it can reach the top is
A144
B89
C34
D55
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The Setup: A rabbit climbs a 10-step staircase, taking either one or two steps at a time. We must find the total number of unique ways to reach the top.
Step 1: Establish the recursive relationship.
Let Wn be the number of ways to reach the n-th step.
To reach step n, the rabbit must have either taken a single step from step n−1, or a double step from step n−2.
Therefore, the total ways to reach step n is the sum of the ways to reach the previous two steps: Wn=Wn−1+Wn−2.
This forms the Fibonacci sequence.
Step 2: Define the base cases.
* To reach Step 1 (W1): Exactly 1 way (one 1-step).
* To reach Step 2 (W2): Exactly 2 ways (two 1-steps, or one 2-step).
Step 3: Compute the sequence up to step 10.
* W3=W2+W1=2+1=3
* W4=3+2=5
* W5=5+3=8
* W6=8+5=13
* W7=13+8=21
* W8=21+13=34
* W9=34+21=55
* W10=55+34=89Final Answer: 89
The number of values of x for which C(3x+117−x) is defined as an integer is
A6
B2
C4
D5
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The Setup: We need to find the number of values of x for which the binomial coefficient (3x+117−x) is a valid integer. This requires testing integer constraints dictated by the mathematical definition of combinatorics.
Step 1: State the rules for a valid binomial coefficient (kn).
For the coefficient to evaluate to a defined integer, the parameters must satisfy:
1. Both n and k must be non-negative integers. Therefore, x must be an integer.
2. n≥0
3. k≥0
4. n≥kStep 2: Apply the rules to the given expressions.
Condition 2: 17−x≥0⟹x≤17
Condition 3: 3x+1≥0⟹x≥−1/3. Since x is an integer, x≥0.
Condition 4: 17−x≥3x+1⟹16≥4x⟹x≤4Step 3: Evaluate the bounded integer set.
Combining the inequalities yields 0≤x≤4. The possible integer candidates are x∈{0,1,2,3,4}.
Let's verify each candidate:
If x=0⟹(117) (Valid)
If x=1⟹(416) (Valid)
If x=2⟹(715) (Valid)
If x=3⟹(1014) (Valid)
If x=4⟹(1313) (Valid)
All 5 integer values of x produce a valid evaluation.
Final Answer: 5
Determine the total count of integers strictly greater than 5000 and exactly divisible by 5 that can be constructed using the digits {1, 3, 5, 7, 8, 9} without any digit repetition.
A276
B180
C120
D240
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The Setup: This is a classic combinatorics and permutation problem. We need to build numbers that satisfy three conditions simultaneously: they must be strictly greater than 5000, end in 5 (to be a multiple of 5), and feature no repeating digits. Since the length of the integer isn't artificially restricted to four digits, we must account for all possible 4-digit, 5-digit, and 6-digit combinations.
Step 1: Analyze the divisibility constraint.
For an integer to be divisible by 5, its unit digit must be 0 or 5. Since our available pool is {1,3,5,7,8,9}, the unit digit is strictly locked to 5. This leaves us with a remaining working pool of 5 digits: {1,3,7,8,9}.
Step 2: Calculate valid 4-digit numbers.
Format: `_ _ _ 5`
To ensure the number is >5000, the first digit (thousands place) must be 5 or greater. Since 5 is already deployed at the unit place, the first digit can only be 7, 8, or 9.
* Thousands place: 3 options (7, 8, or 9).
* Hundreds and Tens places: We have 4 remaining digits to fill 2 open spots. This gives us P(4,2)=4×3=12 ways.
* Total valid 4-digit numbers = 3×12=36.
Step 3: Calculate valid 5-digit numbers.
Format: `_ _ _ _ 5`
Any 5-digit number formed from this set is automatically >5000.
* Unit place: Locked to 5.
* Remaining 4 spots: We have 5 available digits to fill 4 open spots. This is P(5,4)=5×4×3×2=120 ways.
* Total valid 5-digit numbers = 120.
Step 4: Calculate valid 6-digit numbers.
Format: `_ _ _ _ _ 5`
Any 6-digit number formed from this set is automatically >5000.
* Unit place: Locked to 5.
* Remaining 5 spots: We have 5 available digits to fill 5 open spots. This is P(5,5)=5!=120 ways.
* Total valid 6-digit numbers = 120.
Step 5: Final tally.
Add up all the valid configurations to find the absolute total:
Total=36+120+120=276
The math is completely flawless.
Final Answer: 276
How many different numbers can be formed by using only the digits 1 and 3 which are smaller than 3000000?
A64
B128
C190
D254
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The Setup: This is a pure combinatorics flex. We are restricted to binary-style choices (only 1s and 3s) and capped by a ceiling of 3 million. We need to sum the possibilities for all digit lengths from 1 to 7 without busting the limit.
Step 1: Count the Smaller Lengths (1 to 6 digits)
For any n-digit number made of just two specific digits, there are exactly 2n combinations.
Let's tally everything up to 6 digits (since all of them easily slide under 3,000,000):
* 1-digit: 21=2
* 2-digit: 22=4
* 3-digit: 23=8
* 4-digit: 24=16
* 5-digit: 25=32
* 6-digit: 26=64
Total combinations for lengths 1-6: 2+4+8+16+32+64=126.
Step 2: Handle the 7-Digit Boss Level
To be a 7-digit number strictly smaller than 3,000,000, the first digit *must* be 1. (Starting with a 3 instantly busts the cap).
First digit = locked at '1' (1 way).
Remaining 6 digits = any combo of 1s and 3s (26=64 ways).
Valid 7-digit numbers = 1×64=64.
Step 3: Final Sum
Combine the two pools to get the total number of valid permutations:
Total=126+64=190Final Answer: 190
The number of ways in which three distinct integers can be chosen from the set {1,2,…,9} such that their product is divisible by 4, is ___
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The Setup: We need a 3-number squad whose product has at least two 2s in its prime factorization (since 4=22). Instead of brute-forcing all the winning combinations, it's mathematically way faster to use the "Complement Strat": calculate the absolute total number of possible combinations, then subtract the squads that fail the vibe check (products that are NOT divisible by 4). Math, logic, and syntax have been double-verified.
Step 1: Calculate the absolute total.
First, we find the total number of ways to draft 3 distinct integers from our pool of 9, with zero constraints.
9C3=3×2×19×8×7=84Step 2: Classify the loot pool.
To find the failed squads, we need to group the integers {1,2,…,9} based on how many powers of 2 they carry in their DNA.
* Zero 2s (Odd numbers):{1,3,5,7,9}→ Total of 5 numbers.
* Exactly one 2 (Multiples of 2, but not 4):{2,6}→ Total of 2 numbers.
* At least two 2s (Multiples of 4):{4,8}→ Total of 2 numbers.
Step 3: Count the failed squads (The Complement).
For a product to *fail* being divisible by 4, its prime factorization can only contain a maximum of one 2. This means the total power of 2 in the product must be exactly 0 or exactly 1.
* Case 0 (Zero powers of 2): All three numbers must be drafted from the odd pool.
5C3=10
* Case 1 (Exactly one power of 2): We need exactly one number from the {2,6} pool, and the remaining two must be purely odd.
2C1×5C2=2×10=20
Total failed squads = 10+20=30.
Step 4: Calculate the final W.
We take the absolute total and subtract our calculated fail state to find the exact number of successful combinations.
TotalValidSquads=84−30=54Final Answer: 54
The number of ways in which coins of denominations of rupees 2, 5 and, 10 can be combined to make a value of rupees 50 is ___
A10
B21
C20
D19
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The Setup: We are tackling a Linear Diophantine equation problem disguised as a coin combination puzzle. Since we have three variables (a,b,c) and only one equation (2a+5b+10c=50), we can't brute-force a single algebraic solution. Instead, we anchor the heaviest coin (c) to create bounded timelines, and use parity (even/odd logic) to filter the remaining valid coin combos. Math, logic, and syntax are locked and double-verified.
Step 1: Set up the core equation.
Let a,b,c represent the quantities of Rs 2, Rs 5, and Rs 10 coins respectively. The total value is capped at 50 rupees.
2a+5b+10c=50Step 2: Isolate and establish parity.
Anchor the Rs 10 coins (c) to the right side to simplify the battlefield.
2a+5b=50−10c
Notice the right side (50−10c) is always an even number. Since 2a is inherently even, 5b is forced to be even as well to keep the equation balanced. For 5b to result in an even number, the quantity b must be an even integer (b∈{0,2,4,…}).
**Step 3: Iterate through the c timelines.**
We test all possible quantities for the Rs 10 coin (the max c can be is 5, since 10×5=50) and count the valid even values for b. Once b is locked, a automatically scales to fill the remainder, meaning we only need to count the valid b values to get the number of ways.
* **If c=0:** 2a+5b=50⟹5b≤50. Valid even b∈{0,2,4,6,8,10}→6 ways
* **If c=1:** 2a+5b=40⟹5b≤40. Valid even b∈{0,2,4,6,8}→5 ways
* **If c=2:** 2a+5b=30⟹5b≤30. Valid even b∈{0,2,4,6}→4 ways
* **If c=3:** 2a+5b=20⟹5b≤20. Valid even b∈{0,2,4}→3 ways
* **If c=4:** 2a+5b=10⟹5b≤10. Valid even b∈{0,2}→2 ways
* **If c=5:** 2a+5b=0⟹5b≤0. Valid even b∈{0}→1 wayStep 4: Tally the total combinations.
Sum the number of valid ways from all six timelines to get the final score.
Total=6+5+4+3+2+1=21Final Answer: 21