Question:

Let 0rn0 \leq r \leq n. If nCr+1:nCr:nCr1=55:35:21,^nC_{r+1} : ^nC_r : ^nC_{r-1} = 55 : 35 : 21,then 2n+5r2n + 5r is equal to:

Updated On: Mar 20, 2025
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The Correct Option is C

Solution and Explanation

n+1CrnCr=5535 \frac{{n+1}C_r}{nC_r} = \frac{55}{35}

(n+1)!(r+1)!(nr)!÷n!r!(nr)!=117 \frac{(n+1)!}{(r+1)!(n-r)!} \div \frac{n!}{r!(n-r)!} = \frac{11}{7}

n+1r+1=117 \frac{n+1}{r+1} = \frac{11}{7}

7n=4+11r 7n = 4 + 11r

nCrn1Cr1=3521 \frac{nC_r}{n-1C_{r-1}} = \frac{35}{21}

n!r!(nr)!÷(n1)!(r1)!(nr)!=53 \frac{n!}{r!(n-r)!} \div \frac{(n-1)!}{(r-1)!(n-r)!} = \frac{5}{3}

nr=53 \frac{n}{r} = \frac{5}{3}

3n=5r 3n = 5r

Solving for r=6 r = 6 and n=10 n = 10 :

2n+5r=50 2n + 5r = 50

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