If $n$ and $r$ are positive integers such that $r < n$ , then $^{n} C_{r} +^{n} C_{r - 1} =$

^{2 n} C_{2 r - 1}

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^{(n + 1)} C_{r}

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^{n} C_{r + 1}

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^{(n + 1)} C_{r + 1}

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Solution

The correct option is B $^{(n + 1)} C_{r}$
$^{n} C_{r} +^{n} C_{r - 1} = \frac{n!}{(n - r)! r!} + \frac{n!}{(n - r + 1)! (r - 1)!}$
$= \frac{n! (n - r + 1)}{(n - r + 1)! r!} + \frac{n! r}{(n - r + 1)! r!} = \frac{n!}{(n - r + 1)! r!} (n - r + 1 + r)$
$= \frac{(n + 1)!}{({n + 1} - r)! r!} =^{n + 1} C_{r}$

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Similar questions

Let r and n be positive integers such that $1 \leq r \leq n .$ Then prove the following :

(i) $\frac{^{n} C_{r}}{^{n} C_{r - 1}} = \frac{n - r + 1}{r}$
(ii) $n^{n - 1} C_{r - 1} = (n - r + 1)^{n} C_{r - 1}$
(iii) $\frac{^{n} C_{r}}{^{n - 1} C_{r - 1}} = \frac{n}{r}$
(iv) $^{n} C_{r} + 2^{n} C_{r - 1} +^{n} C_{r - 2} =^{n + 2} C_{r}$

Q. Let r and n be positive integers such that 1 ≤ r ≤ n. Then prove the following:
(a)

\frac{{}^{n}C_{r}}{{}^{n}C_{r - 1}} = \frac{n - r + 1}{r}

(b) n ·_{^{n −}}_¹C_r_{− 1} = (n − r + 1) ⁿC_r_{− 1}
(c)

\frac{{}^{n}C_{r}}{{}^{n - 1}C_{r - 1}} = \frac{n}{r}

(iv) ⁿC_r + 2 · ⁿC_r_{− 1} + ⁿC_r_{− 2} = ^{n +}²C_r.

Q. If

(2 \leq r \leq n),

then

^{n} C_{r} + 2 \cdot^{n} C_{r + 1} +^{n} C_{r + 2}

is equal to

Q. If n and r are two positive integers such that

n \geq r

, then

^{n} C_{r + 1} +^{n} C_{r} =

$^{n} C_{r} +^{n} C_{r - 1}$ is equal to

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If n and r are positive integers such that r<n, then nCr+nCr−1=

The correct option is B (n+1)CrnCr+nCr−1=n!(n−r)!r!+n!(n−r+1)!(r−1)!=n!(n−r+1)(n−r+1)!r!+n!r(n−r+1)!r!=n!(n−r+1)!r!(n−r+1+r)=(n+1)!({n+1}−r)!r!=n+1Cr

If $n$ and $r$ are positive integers such that $r < n$ , then $^{n} C_{r} +^{n} C_{r - 1} =$