Question

n and r, if npr =120 ,ncr=20​

Answer 1

Answer:

The Value of r is 3

The value of n is 6

Step-by-step explanation:

ncr = npr ÷ r!

20 = 120 ÷ r!

r! = 6

so r =3

Answer 2

The values are : n = 6 , r = 3

Given :

$\displaystyle \sf{ {}^{n}P_{r} = 120 \: }$

$\displaystyle \sf{ {}^{n}C_{r} = 20 \: }$

To find :

The value of n and r

Solution :

Step 1 of 3 :

Write down the given data

Here it is given that

$\displaystyle \sf{ {}^{n}P_{r} = 120 \: }$

$\displaystyle \sf{ {}^{n}C_{r} = 20 \: }$

Step 2 of 3 :

Calculate the value of r

$\displaystyle \sf{ {}^{n}P_{r} = 120 \: }$

$\displaystyle \sf{ {}^{n}C_{r} = 20 \: }$

$\implies \displaystyle \sf{ \frac{n! }{(n - r)! } = 120\: } \: . \: . \:. \:. \:. \:(1)$

Again$\displaystyle \sf{ {}^{n}C_{r} = 20 \: }$

$\implies \displaystyle \sf{ \frac{n! }{r!(n - r)! } = 20\: } \: . \: . \:. \:. \:. \:(2)$

Equation (1) ÷ Equation (2) gives

$\sf{r ! = 6}$

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

$\implies \sf{r ! = 3 !}$

$\implies \sf{r = 3 \: }$

Step 3 of 3 :

Calculate the value of n

From Equation (2) we get

$\displaystyle \sf{ \frac{n! }{3!(n - 3)! } = 20\: } \: \: \:$

$\displaystyle \sf{ \implies n(n - 1)(n - 2) = 20 \times 3!}$

$\displaystyle \sf{ \implies n(n - 1)(n - 2) = 20 \times 6}$

$\displaystyle \sf{ \implies n(n - 1)(n - 2) = 120 }$

$\displaystyle \sf{ \implies n(n - 1)(n - 2) = 6 \times 5 \times 4}$

$\displaystyle \sf{ \implies n(n - 1)(n - 2) = 6 \times (6 - 1) \times (6 - 2)}$

Comparing both sides we get n = 6

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