Question

69. Find n, if nPr=840 and nCr=35.​

Answer 1

ɧɛყ ɬɧɛཞɛ❤

1) nPr =n! /(n-r)!

2) nCr = n! /[(n-r)! r! ]

3)r! =(nPr) /nCr

it is given that,

nPr = 840..... (1)

nCr =35...(2)

r! =(nPr) /(nCr)

=840/35

=24

r! = 4!

therefore.,

r=4

substitute r=value in eq(1),

we get

n! /(n-4) ! =840

n(n-1) (n-2) (n-3) = 840

n(n-1) (n-2) (n-3) = 7×6×5×4

therefore,

n=7✔️✔️

✌✌✌✌✌✌✌✌

Answer 2

The value of n is 7.

Explanation:

Since , the number of permutations of selecting r things from n = $^nP_r=\dfrac{n!}{(n-r)!}$

The number of combinations of selecting r things from n = $^nC_r=\dfrac{n!}{r!(n-r)!}$

Given : $^nP_r=840$

$^nC_r=35$

⇒ $\dfrac{n!}{(n-r)!}=840\ \ \ ,\ \dfrac{n!}{r!{(n-r)}!}=35$

⇒ $\dfrac{840}{r!}=35$

⇒ $r!=\dfrac{840}{35}=24=4!$

i.e. r= 4

⇒  $^nP_4=\dfrac{n!}{(n-4)!}=840=7\times6\times5\times4=\dfrac{7\times6\times5\times4\times3\times2\times1}{3\times2\times1}$

⇒$\dfrac{n!}{(n-4)!}=\dfrac{7!}{3!}=\dfrac{7!}{(7-3)!}$

⇒ n= 7

such that $^7P_4=840\ \ \& \ \ ^7C4=35$

Hence, the value of n is 7.

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