Prove that nCrn-1Cr-1-nr when 1- r - n-

#10; ^ n C_ r ^ n C_ n r = n r LHS= ^ n #10;C_ r ^ n C_ n r ⇒ n! (n r)!r! #10; #160; (n 1)! (n r)!(r 1)! #160; = n r #10; ...

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Standard XII

Mathematics

Combination

Prove that ...

Question

Prove that $\frac{^{n} C_{r}}{^{n - 1} C_{r - 1}} = \frac{n}{r}$ when $1 \leq r \leq n$ .

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Solution

$\frac{^{n} C_{r}}{^{n} C_{n - r}} = \frac{n}{r} L H S = \frac{^{n} C_{r}}{^{n} C_{n - r}} \Rightarrow \frac{\frac{n!}{(n - r)! r!}}{\frac{(n - 1)!}{(n - r)! (r - 1)!}} = \frac{n}{r}$

Hence proved as $L H S = R H S$

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Similar questions

Q. Prove that

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

where

1 \leq r \leq n

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. Prove that

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

Q. Prove that

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

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.

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Prove that nCrn−1Cr−1=nr when 1≤r≤n.

nCrnCn−r=nrLHS=nCrnCn−r⇒n!(n−r)!r!(n−1)!(n−r)!(r−1)!=nr Hence proved as LHS=RHS

Prove that $\frac{^{n} C_{r}}{^{n - 1} C_{r - 1}} = \frac{n}{r}$ when $1 \leq r \leq n$ .

$\frac{^{n} C_{r}}{^{n} C_{n - r}} = \frac{n}{r} L H S = \frac{^{n} C_{r}}{^{n} C_{n - r}} \Rightarrow \frac{\frac{n!}{(n - r)! r!}}{\frac{(n - 1)!}{(n - r)! (r - 1)!}} = \frac{n}{r}$

Hence proved as $L H S = R H S$