Prove that-i n -n-r -nn-1n-2 -n-r-1ii n -n-r - r -n -n-r-1 -r-1 -n-1 - r -n-r-1 -

(i) #160;LHS= #160;n!(n r)! #160; #160; #160; #160; #160; #160; #160; #160; #160; #160; #160; #160; #160;=nn 1n 2n 3n 4 #160;... ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

Mathematics

Absolute Value Function

Prove that:i ...

Question

Prove that:
(i) $\frac{n!}{(n - r)!}$ = n (n − 1) (n − 2) ... (n − (r − 1))

(ii) $\frac{n!}{(n - r)! r!} + \frac{n!}{(n - r + 1)! (r - 1)!} = \frac{(n + 1)!}{r! (n - r + 1)!}$

Open in App

Solution

$(i) LHS = \frac{n!}{(n - r)!} = \frac{n (n - 1) (n - 2) (n - 3) (n - 4) . . . (n - r + 1) [(n - r)!]}{(n - r)!} = n (n - 1) (n - 2) (n - 3) (n - 4) . . . (n - r + 1) = n (n - 1) (n - 2) (n - 3) (n - 4) . . . [n - (r - 1)] = RHS$

$(i i) LHS = \frac{n!}{(n - r)! r!} + \frac{n!}{(n - r + 1)!} = \frac{n!}{(n - r)! r!} + \frac{n!}{(n - r + 1) [(n - r)!]} = \frac{n! (n - r + 1) + n! r!}{r! (n - r + 1) [(n - r)!]} = \frac{n! (n + 1) - n! r! + n! r!}{r! (n - r + 1) (n - r)!} = \frac{n! (n + 1)}{r! (n - r + 1) (n - r)!} = \frac{(n + 1!)}{r! (n - r + 1)!} = RHS Hence proved .$

Suggest Corrections

Similar questions

Prove that:
(i) $\frac{n!}{(n - r)!}$
= n(n-1)(n-2)...(n-(r-1))
(ii) $\frac{n!}{(n - r)! r!} + \frac{n!}{(n - r + 1)! (r - 1)!}$
= $\frac{(n + 1)!}{r! (n - r + 1)!}$

Q. Prove that:

\frac{n!}{(n - r)!} = n (n - 1) (n - 2) . . . (n - (r - 1))

Prove that:

n! / r! x (n-r)! + n! / (r-1)! x (n-r+1) = (n+1)! / r! x (n-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.

Q. If

(1 - x)^{- n} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . . . . + a_{r} x^{r} + . . . . .

, then

a_{0} + a_{1} + a_{2} + . . . . . . + a_{r}

is equal to

Join BYJU'S Learning Program

Prove that: (i) n!(n-r)! = n (n − 1) (n − 2) ... (n − (r − 1)) (ii) n!(n-r)! r!+n!(n-r+1)! (r-1)!=(n+1)!r! (n-r+1)!

(i) LHS= n!(n-r)! =nn-1n-2n-3n-4 ... n-r+1n-r!(n-r)! =nn-1n-2n-3n-4 ... n-r+1 =nn-1n-2n-3n-4 ... n-r-1 = RHS (ii) LHS = n!n-r!r!+n!n-r+1! =n!n-r!r!+ n!(n-r+1) [(n-r)!] =n!n-r+1+ n!r!r!n-r+1 [(n-r)!] =n!n+1-n!r! + n!r!r!n-r+1n-r! =n!(n+1)r!n-r+1n-r! =n+1!r!n-r+1! = RHS Hence proved.

Prove that:
(i) $\frac{n!}{(n - r)!}$ = n (n − 1) (n − 2) ... (n − (r − 1))

(ii) $\frac{n!}{(n - r)! r!} + \frac{n!}{(n - r + 1)! (r - 1)!} = \frac{(n + 1)!}{r! (n - r + 1)!}$