If n and r are integers such that 1 - r - n- then n-n-1Cr - 1 -

The correct option is D r^nC_r n.^n 1C_r 1 = n.(n 1)!(n r)!(r 1)! =r.n!(n r)!r! =r. ^nC_r ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Pascal Triangle

If n and ...

Question

If $n$ and $r$ are integers such that $1 \leq r \leq n$ , then $n .^{n - 1} C_{r - 1} =$

^{n} C_{r}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

n^{n} C_{r}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

r^{n} C_{r}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

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

Suggest Corrections

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.

If $1 \leq r \leq n$ then $n^{n - 1} C_{r - 1}$ is equal to .........

Q. If

n

and

r

are positive integers such that

r < n

, then

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

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

n \geq r

, then

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

Join BYJU'S Learning Program

If n and r are integers such that 1≤r≤n, then n.n−1Cr−1=

The correct option is D rnCrn.n−1Cr−1=n.(n−1)!(n−r)!(r−1)!=r.n!(n−r)!r!=r.nCr

If $n$ and $r$ are integers such that $1 \leq r \leq n$ , then $n .^{n - 1} C_{r - 1} =$

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