Question

If six times the number of permutations of n things taken 3 at a time is equal to seven times the number of permutations of (n - 1) things taken 3 at a time, find n.

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Answer 1

ANSWER:

Given:

• 6 times the number of permutations of n things taken 3 at a time = 7 times the number of permutations of (n - 1) things taken 3 at a time

To Find:

• Value of n

Solution:

$\text{We know that,}\\\\:\hookrightarrow\text{Number of permutations of n things taken r at a time = ^nP_r }\\\\\text{So,}\\\\:\implies\text{Number of permutations of n things taken 3 at a time = ^nP_3 }\\\\\text{And,}\\\\:\implies\text{Number of permutations of (n-1) things taken 3 at a time = ^{n-1}P_3 }\\\\\text{We are given that,}\\\\:\implies 6\times ^nP_3=7\times ^{n-1}P_3\\\\\text{We know that,}\\\\:\hookrightarrow ^nP_r=\dfrac{(n)!}{(n-r)!}$

$\text{So,}\\\\:\implies6\times\left(\dfrac{(n)!}{(n-3)!}\right)=7\times\left(\dfrac{(n-1)!}{(n-1-3)!}\right)\\\\:\implies6\times\left(\dfrac{(n)!}{(n-3)!}\right)=7\times\left(\dfrac{(n-1)!}{(n-4)!}\right)\\\\:\implies6\times\left(\dfrac{(n)(n-1)(n-2)(n-3)!}{(n-3)!}\right)=7\times\left(\dfrac{(n-1)(n-2)(n-3)(n-4)!}{(n-4)!}\right)\\\\\text{(n-3)! and (n-4)! get cut in LHS and RHS respectively. So,}\\\\:\implies6(n)(n-1)(n-2)=7(n-1)(n-2)(n-3)\\\\\text{(n-1)(n-2) gets cancelled on both sides of equation.}\\\\\text{So,}\\\\:\implies6(n)=7(n-3)\\\\:\implies6n=7n-21\\\\:\implies7n-6n=21\\\\\bf{:\implies n=21}\\\\\text{\bf{The value of n is 21.}}$

Formula Used:

• ^(n) P _(r) = (n)!/(n-r)!

Answer 2

best answer ke liye refer the attachment.

Agar achha lage to like zarur krna.