Question

If nP100 = nP99 then find the value of n (Permutation Combination)

Answer 1

make me brainlist ....

Answer 2

$nPr = \frac{n!}{(n - r)!}$

Here,

$nP100 = nP99 \\ \\ = \frac{n!}{(n - 100)!} = \frac{n!}{(n - 99)!} \\ \\ n!(n - 99)! = n!(n - 100)! \\ \\ (n - 99)! = (n - 100)!$

We know that 0! = 1! = 1.

Comparing with this, we can make an equation like,

$(n - 99)! = (n - 100)! = 1! = 0! = 1 \\ \\ \therefore n = 100$

I can't make an actual proof.

But, here, through this question and my solution, I can understand one thing.

If nPr = nP(r - 1), then n = r.

Hope this may be helpful.

Please mark my answer as the brainliest if this helps you.

Thank you. Have a nice day.