Question

If nPr = 3024 and nCr = 126 then find n and r.

Answer 1

SOLUTION

GIVEN

$\displaystyle \sf{ {}^{n}P_{r} = 3024 \: }$

$\displaystyle \sf{ {}^{n}C_{r} = 126 \: }$

TO DETERMINE

The value of n and r

EVALUATION

Here it is given that

$\displaystyle \sf{ {}^{n}P_{r} = 3024 \: }$

$\implies \displaystyle \sf{ \frac{n! }{(n - r)! } = 3024\: } \: \: \: .....(1)$

Again

$\displaystyle \sf{ {}^{n}C_{r} = 126 \: }$

$\implies \displaystyle \sf{ \frac{n! }{r!(n - r)! } = 126\: } \: \: \: .....(2)$

Equation (1) ÷ Equation (2) gives

$\sf{r ! = 24}$

$\implies \sf{r ! = 4 \times 3 \times 2 \times 1}$

$\implies \sf{r ! = 4 !}$

$\implies \sf{r = 4 \: }$

From Equation (2) we get

$\displaystyle \sf{ \frac{n! }{4!(n - 4)! } = 126\: } \: \: \:$

$\implies \displaystyle \sf{ \frac{n(n - 1)(n - 2)(n - 3)}{4!} = 9 \times 7 \times 2\: } \: \: \:$

$\implies \displaystyle \sf{ {n(n - 1)(n - 2)(n - 3)} = {4!} \times 9 \times 7 \times 2\: } \: \: \:$

$\implies \displaystyle \sf{ {n(n - 1)(n - 2)(n - 3)} = 24 \times 9 \times 7 \times 2\: } \: \: \:$

$\implies \sf{ n(n - 1)(n - 2)(n - 3) = 9 \times 8 \times 7 \times 6\: }$

$\implies \sf{ n(n - 1)(n - 2)(n - 3) = 9 \times (9 - 1) \times (9 - 2) \times (9 - 3)\: }$

Comparing both sides we get

$\sf{n = 9}$

FINAL ANSWER

The required value of n and r 9 and 4 respectively

━━━━━━━━━━━━━━━━

Learn more from Brainly :-

1. (32C0)^2-(32C1)^2+(32C2)^2-..........(32C32)^2=

https://brainly.in/question/28583245

2. The value of 20 sigma 50-r C6 is equal to

https://brainly.in/question/25207335