Question

If nPr=1680 and nCr=70. find n and r.

Answer 1

$\huge{\boxed{\texttt{Answer :-}}}$

♦ As given in the question

>> $^{n}P_{r} = 1680$

>> $^{n}C_{r} = 70$

♦ Before solving we must know ,

• What is Permutations ?

>> Number of ways in which we arrange the given number of sets is called Permutation.

• What is Combination ?

>> Number of distinct ways from which we can select the set is called Combination.

• How to calculate Permutation ?

>> $^{n}P_{r} = \dfrac{n!}{(n-r)!}$

• How to calculate Combination ?

>> $^{n}C_{r} = \dfrac{n!}{r! \times (n-r)!}$

♦ By looking at the formula for $^{n}P_{r} \: and \: ^{n}C_{r}$ we can say

$^{n}C_{r} = \dfrac{^{n}P_{r}}{r!}$

♦ Then for "r" by using above

$\dfrac{1680}{ r!} = 70$

$\implies 1680 = 70(r!)$

$\implies \dfrac{1680}{ 70} = r!$

$\implies 24 = r!$

$\implies r! = (1)(2)(3)(4)$

$\implies r = 4$

♦ Now for "n"

$\dfrac{n!}{(n-r)!} = 1680$

$\implies \dfrac{n!}{(n-4)!} = 1680$

$\implies \dfrac{n(n-1)(n-2)(n-3)(n-4)!}{(n-4)!} = 1680$

$\implies n(n-1)(n-2)(n-3) = 1680$

♦ By prime using factorization of 1680

1680 = $2^4 \times 3 \times 5 \times 7$

So, we can write

1680 = 8 × 7 × 6 × 5

♦ Then

$\implies n(n-1)(n-2)(n-3) = (8)(7)(6)(5)$

>> By comparing RHS to LHS

We get

$\textbf{n = 4 }$

♦ So the values of

$\bold{ r = 4}$

$\bold{n = 8 }$