Question

Find X where 1\4! + 3\6! = X\8!

Answer 1

Question :-

Find x -

$\sf \dfrac{1}{4!} + \dfrac{3}{6!} = \dfrac{x}{8!}$

Answer :-

Factorial of a number n is the product of all positive integers less than or equal to n. It is denoted by n!.

For example :- 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1

$\implies\sf \dfrac{1}{4!} + \dfrac{3}{6!} = \dfrac{x}{8!}$

$\implies\sf \dfrac{1}{4 \times 3 \times 2\times 1} + \dfrac{\not3}{6 \times 5 \times 4 \times \not3 \times 2\times 1} = \dfrac{x}{8!}$

$\implies\sf \dfrac{1}{24} + \dfrac{1}{240} = \dfrac{x}{8!}$

$\implies\sf \dfrac{10}{240} +\dfrac{1}{240} = \dfrac{x}{8!}$

$\implies\sf \dfrac{11}{240} = \dfrac{x}{8!}$

$\implies\sf \dfrac{11}{240} = \dfrac{x}{8\times 7 \times 6 \times 5 \times 4 \times 3 \times 2\times 1}$

$\implies\sf \dfrac{11}{240} = \dfrac{x}{40320}$

$\implies\sf 240 \times x = 40320 \times 11$

$\implies\sf 240x = 443520$

$\implies\sf x = 1848$

Value of x = 1848