Question

If $\left( {7!} \right)!$ is divisible by ${\left( {7!} \right)^{k!}}\left( {6!} \right)$ the $k$ can be
A. 2
B. 3
C. 5
D. 6

Answer 1

Expand the factorial of 7 and 6 using the property $n! = n.\left( {n - 1} \right)....3.2.1$ and rewrite the given expression. Then, use hit and trial method to find the what value of $k$ will divide $\left( {7!} \right)!$ by ${\left( {7!} \right)^{k!}}\left( {6!} \right)$.

Complete step by step solution:
We know that $n! = n.\left( {n - 1} \right)....3.2.1$
Then, we have $7! = 7.6.5.4.3.2.1 = 5040$
And $6! = 6.5.4.3.2.1 = 720$
Then, we are given that $\left( {5040} \right)!$ is divisible by ${\left( {5040} \right)^{k!}}\left( {720} \right)$
That is,
$\dfrac{{\left( {5040} \right)!}}{{{{\left( {5040} \right)}^{k!}}\left( {720} \right)}}$
Now, we have to find the value of $k$ such that $\left( {7!} \right)!$ is divisible by ${\left( {7!} \right)^{k!}}\left( {6!} \right)$
We will find the value of $k$ by hit and trial method.
Let $k = 2$
$\dfrac{{\left( {5040} \right)!}}{{{{\left( {5040} \right)}^{2!}}\left( {720} \right)}}$
Now, $2! = 2.1 = 2$
$\dfrac{{\left( {5040} \right) \times \left( {5039} \right) \times \left( {5038} \right).....\left( {721} \right) \times \left( {720} \right)!}}{{\left( {5040} \right) \times \left( {5040} \right) \times \left( {720} \right)!}}$
Which can be simplified as
$\dfrac{{\left( {5039} \right) \times \left( {5038} \right).....\left( {721} \right)}}{{\left( {5040} \right)}}$
Clearly, 5040 will divide the numerator completely as $5040 = 1260 \times 4$

Hence, the value of $k$ can be 2.
Thus, option A is correct.

Note:
A factorial is a number that multiplies the number by every other number that is less than it. The expansion of $n!$ is $n.\left( {n - 1} \right)....3.2.1$. It helps in counting permutations and combinations. Also, we say a number has completely divided when no remainder is left after division.