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Permutations and Combinations question

2024 · 6 Apr · Shift 2 · Q47
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  5. /2024 · 6 Apr · Shift 2 · Q47

Permutations and Combinations question

2024 · 6 Apr · Shift 2 · Q47

JEE MainMathematicsPermutations and CombinationsMCQ+4 / −1
Let 0≤r≤n0 \leq r \leq n0≤r≤n. If n+1Cr+1:nCr:n−1Cr−1=55:35:21{ }^{n+1} C_{r+1}:{ }^n C_r:{ }^{n-1} C_{r-1}=55: 35: 21n+1Cr+1​:nCr​:n−1Cr−1​=55:35:21, then 2n+5r2 n+5 r2n+5r is equal to :
  1. A
    62
  2. B
    60
  3. C
    55
  4. D
    50
View written solutionFree

Correct answer: D

  1. Let A=n+1Cr+1,B=nCr,C=n−1Cr−1A={}^{n+1}C_{r+1},\quad B={}^{n}C_r,\quad C={}^{n-1}C_{r-1}A=n+1Cr+1​,B=nCr​,C=n−1Cr−1​ and given A:B:C=55:35:21.A:B:C=55:35:21.A:B:C=55:35:21.

So, AB=5535=117,BC=3521=53.\frac{A}{B}=\frac{55}{35}=\frac{11}{7}, \qquad \frac{B}{C}=\frac{35}{21}=\frac{5}{3}.BA​=3555​=711​,CB​=2135​=35​.

  1. Use the binomial coefficient ratio formulas.

First,

\frac{\frac{(n+1)!}{(r+1)!(n-r)!}}{\frac{n!}{r!(n-r)!}} =\frac{n+1}{r+1}.$$ Hence, $$\frac{n+1}{r+1}=\frac{11}{7}.$$ So, $$7(n+1)=11(r+1)$$ $$7n+7=11r+11$$ $$7n-11r=4. \quad (1)$$ 3. Next, $$\frac{{}^nC_r}{{}^{n-1}C_{r-1}}= \frac{\frac{n!}{r!(n-r)!}}{\frac{(n-1)!}{(r-1)!(n-r)!}} =\frac{n}{r}.$$ Hence, $$\frac{n}{r}=\frac{5}{3}.$$ So, $$3n=5r. \quad (2)$$ 4. Solve the two equations. From (2), $$n=\frac{5r}{3}.$$ Substitute into (1): $$7\left(\frac{5r}{3}\right)-11r=4$$ $$\frac{35r}{3}-\frac{33r}{3}=4$$ $$\frac{2r}{3}=4$$ $$2r=12$$ $$r=6.$$ Then, $$n=\frac{5}{3}\cdot 6=10.$$ 5. Now compute $$2n+5r=2(10)+5(6)=20+30=50.$$ 6. Checking options: - A: 62 - B: 60 - C: 55 - D: 50 So the correct option is **D**.
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