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Permutations and Combinations question

2003 · Shift 0 · Q108

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JEE Main
Mathematics
Permutations and Combinations
2003 · Shift 0 · Q108

Permutations and Combinations question

2003 · Shift 0 · Q108

JEE MainMathematicsPermutations and CombinationsMCQ+4 / −1

If

{}^n{C_r}

denotes the number of combination of n things taken r at a time, then the expression

\,{}^n{C_{r + 1}} + {}^n{C_{r - 1}} + 2\, \times \,{}^n{C_r}

equals

A
$\,{}^{n + 1}{C_{r + 1}}$
B
${}^{n + 2}{C_r}$
C
${}^{n + 2}{C_{r + 1}}$
D
$\,{}^{n + 1}{C_r}$

View written solutionFree

Correct answer: C

We need to simplify

{}^nC_{r+1}+{}^nC_{r-1}+2{}^nC_r.

Use Pascal's identity:

{}^nC_k+{}^nC_{k-1}={} ^{n+1}C_k.

First combine

{}^nC_{r+1}+{}^nC_r = {}^{n+1}C_{r+1}.

Next combine

{}^nC_{r-1}+{}^nC_r = {}^{n+1}C_r.

So the given expression becomes

{}^{n+1}C_{r+1}+{}^{n+1}C_r.

Apply Pascal's identity again:

{}^{n+1}C_{r+1}+{}^{n+1}C_r = {}^{n+2}C_{r+1}.

Therefore,

{}^nC_{r+1}+{}^nC_{r-1}+2{}^nC_r = {}^{n+2}C_{r+1}.

Checking options:

A: ${}^{n+1}C_{r+1}$ — incorrect
B: ${}^{n+2}C_r$ — incorrect
C: ${}^{n+2}C_{r+1}$ — correct
D: ${}^{n+1}C_r$ — incorrect

Hence the correct answer is Option C.

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