Question

what is the value of n, if {(n+4)!/(n+2)!} = 72​

Answer 1

Answer:

Step-by-step explanation:

Therefore,

$\frac{1*2*3*....*(n+2)*(n+3)*(n+4)}{1*2*3*...(n+2)} =72\\Therefore ,\\(n+3)*(n+4)=72\\Therefore,\\n^2+3n+4n+7=72\\,n^2+7n=65\\Therefore\\n^2+7n-65=0$

Multiplying both sides by 2,

$2n^2+14n-120=0\\since\\120=20*6\\Therefore,\\2n^2+20n-6n-120=0\\Therfore\\2n^2-6n+20n-120=0\\Therefore\\2n(n-3)+20(n-6)=0$

it should have gotten factorized here but i have done a mistake

anyway this is the method with which you can do it hope it helps

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