Question

solve nC3 = n-1C2 + 84
Ch-Permutation and combination class 11 maths
Please don’t attempt the question if you don’t know the answer…

Answer 1

Step-by-step explanation:

$\implies ^nC_r =^{n-1}C_2+84$

Apply formula $^nC_r =\dfrac{n!}{r!(n-r)!}$

${ \implies \dfrac{n!}{3!(n-3)!} =\dfrac{(n-1)!}{2!(n-1-2)!} + 84 }$

${ \implies \dfrac{n!}{3!(n-3)!} - \dfrac{(n-1)!}{2!(n-1-2)!} = 84 }$

We know that n! = n (n-1)!

${ \implies \dfrac{n(n - 1)!}{3!(n-3)!} - \dfrac{(n-1)!}{2!(n-3)!} = 84 }$

${ \implies \dfrac{(n - 1)!}{(n-3)!} . \bigg( \dfrac{n}{3!} - \dfrac{1}{2!} \bigg)= 84 }$

${ \implies \dfrac{(n - 1)!}{(n-3)!} . \bigg( \dfrac{n}{3 \times 2 \times 1} - \dfrac{1}{2 \times 1} \bigg)= 84 }$

${ \implies \dfrac{(n - 1)!}{(n-3)!} . \bigg( \dfrac{n}{6} - \dfrac{1}{2} \bigg)= 84 }$

${ \implies \dfrac{(n - 1)!}{(n-3)!} . \bigg( \dfrac{n - 3}{6} \bigg)= 84 }$

We know that (n-1)! = (n-1) . (n-2) . (n-3)!

${ \implies \dfrac{(n - 1)(n - 2)(n - 3)!}{(n-3)!} . \bigg( \dfrac{n - 3}{6} \bigg)= 84 }$

${ \implies \dfrac{(n - 1)(n - 2) \cancel{(n - 3)!}}{ \cancel{(n-3)!}} . \bigg( \dfrac{n - 3}{6} \bigg)= 84 }$

${ \implies \dfrac{(n - 1)(n - 2)}{ 1} . \bigg( \dfrac{n - 3}{6} \bigg)= 84 }$

${ \implies \dfrac{(n - 1)(n - 2)(n - 3)}{6} = 84 }$

Now try to make LHS and RHS equal

${ \implies \dfrac{(n - 1)(n - 2)(n - 3)}{6} = 84 \times \dfrac{6}{6} }$

${ \implies \dfrac{(n - 1)(n - 2)(n - 3)}{6} = \dfrac{504}{6} }$

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

${ \implies \dfrac{(n - 1)(n - 2)(n - 3)}{6} = \dfrac{(10 - 1)(10 - 2)(10 - 3)}{6} }$

${ \implies \dfrac{(n - 1)(n - 2)(n - 3)}{ \not6} = \dfrac{(10 - 1)(10 - 2)(10 - 3)}{ \not6} }$

${ \implies (n - 1)(n - 2)(n - 3) = (10 - 1)(10 - 2)(10 - 3)}$

By comparing LHS and RHS, n = 10.

So the value of n is 10.

$\rule{280}{1.7}$

Remember that:

n! called n factorial is the expansion of n × (n-1) × (n-2) × (n-3) . . . 3 × 2 × 1