Question

If the factorial of 10 is written as follows
 $10!=2^p{.3}^q{.5}^r{.7}^s,$
then which of the following statements is true?
(a). $2q=p$
(b). $pqrs=64$
(c). Number of divisors of $10!$ is 280
(d). Number of ways of putting $10!$ as a product of two natural numbers is 135.

Answer 1

Hint: Here we need the explanation on $10!$ . We can find the exponent of a prime number in any factorial by just one single formula. Here in the given question there are all prime numbers only. So, we have no problem. Now by using that formula we can find values of all 4 variables. Now multiply them all together to get their product. Find the interrelation between the first two variables. For the number of divisors also we have a formula let p be a prime number. $E_p\left(n\right)$ is an exponent of p in $n!$.

$E_p\left(n\right)=\left[{\displaystyle \frac{n}{p}}\right]+\left[{\displaystyle \frac{n}{p^2}}\right]+...............\left[{\displaystyle \frac{n}{p^{k-1}}}\right]$ , where $p^k>n$.
If number is written as $2^p{.3}^q{.5}^r{.7}^s.........$
Number of divisors is written as $(r\ne 1)(q\ne 1)(s\ne 1)............$

Complete step-by-step solution -
Factorial:- In mathematics, factorial is an operation, denoted by “$(!)$ “. It represents the product of all numbers between 1 and a given number.
In simple words factorial of a number can be found by multiplying all terms you get by subtracting 1 from a given number repeatedly till you get the number difference as 1. Its representation can be written as: $n!=n\times (n-1)\times (n-2)...................\times 1$
For example: $5!=5\times 4\times 3\times 2\times 1=120.$
Note that we assume $0!=1$ . It is standard value. It has a wide range of applications in combinations..
Given representation of factorial in the question, is written as:
$10!=2^p{3}^q{5}^r{7}^s$ .
By basic knowledge of factorial, we know the formula of exponent.
Let a be prime number $E_p\left(n\right)$ is exponent of p in $$n!$$ ,
$E_a\left(n\right)=\left[{\displaystyle \frac{n}{a}}\right]+\left[{\displaystyle \frac{n}{a^2}}\right]+...............\left[{\displaystyle \frac{n}{a^{k-1}}}\right]$ . where $a^k>n$ .
If we take $a=2,n=10$ , in this formula, we get it as:
$p=\left[{\displaystyle \frac{10}{2}}\right]+\left[{\displaystyle \frac{10}{2^2}}\right]+\left[{\displaystyle \frac{10}{2^3}}\right]$
By simplifying the above equation, we can write it as:
$p=\left[{\displaystyle \frac{10}{2}}\right]+\left[{\displaystyle \frac{10}{4}}\right]+\left[{\displaystyle \frac{10}{8}}\right]=5+2+1$
By simplifying it more, we can write value of p as:
$p=8$
If we take $a=3$ , $n=10$ , in the formula, we get it as:
$q=\left[{\displaystyle \frac{10}{3}}\right]+\left[{\displaystyle \frac{10}{3^2}}\right]$ .
By simplifying the above equation, we get value of q as:
$q=4$ .
If we take $a=5,n=10$ in the formula, we get r as:
$r=\left[{\displaystyle \frac{10}{5}}\right]=2$ .
If we take $a=7,n=10$ in the formula, we get s as:
$s=\left[{\displaystyle \frac{10}{7}}\right]=1$ .
So, by substituting all values we get $10!=2^8\times 3^4\times 5^2\times 7..........$
Option 1:- $2q=p$ , Here $2\left(4\right)=8$ . so this option is correct.
Option 2:- $pqrs=64$ , Here $\left(8\right)\left(4\right)\left(2\right)\left(1\right)=64$ . So, this option is correct.
Option 3:- Number of divisors $=(p+1)(q+1)(r+1)(s+1)$
$=\left(9\right)\left(5\right)\left(3\right)\left(1\right)=270$ .
S, it is wrong.
Option 4:- If there are n divisors we can write them in pairs of 2 in ${\displaystyle \frac{n}{2}}$ ways. Here we have 270. So, 135 us correct.
Therefore, option (1), (2), (4) are correct

Note: Be careful while calculating power of 2. Generally students forgot to take the $2^3$ term. While finding a number of ways of pairs look carefully that repeating must not come because in product $ab=ba$ . So, you must divide by 2. Alternate way is to expand $10!$ directly only by prime numbers.