Solve it:
$N=36!\times 20$ . The difference between the exponent of 3 and exponent of 5 in N is?

Hint: In this question we need to find the difference of two exponents in N. For this, we will find the number of multiples of 3 then count the total power of 3 in those multiples. This will give us an exponent of 3. In the same way, we will then find the exponent of 5 in N. After obtaining these two, we will subtract them and hence we will get our answer.

Complete step-by-step solution
Now, we have to find the difference between the exponents of 3 and 5 in N.
We have been given that $N=36!\times 20$.
On expanding the factorial we will get N as:
$\begin{align}
  & N=36!\times 20 \\
 & \Rightarrow N=36\times 35\times 34\times 33\times .....\times 1\times 20 \\
\end{align}$
Now, we will first find the exponent of 3 in N.
For this, we will count the multiples of 3 in N and count the powers from that.
Now, the multiplies of 3 in the numbers which are multiplied together to form N are:
$3,6,9,12,15,18,21,24,27,30,33,36$
Now, among these multiples of 3, the number of multiplies of 3 which contain only one power of 3, i.e. which are not multiples of 9 and 27 i.e ${{3}^{2}},{{3}^{3}}$ are:
$3,6,12,15,21,24,30,33$
Thus, there are a total of 8 multiples of 3 which have only one power of 3.
Since all these numbers are multiplied to form N, the total exponent of 3 of these numbers will be contributed to the exponent of 3 in N. Since all these multiplies contain 3 as a factor only once, the number of these digits will be the required contribution.
Thus, from here the exponent of 3 is 8. .....(i)
Now, among these multiples of 3, the number of multiples of 3 which contain only the second power of 3, i.e ${{3}^{2}}=9$ which are not the multiples of 27 are:
$9,18,36$
Thus, there are a total of 3 multiples of 3 which contain only the second power of 3.
So, the contribution for the exponent of 3 in N from these numbers will be:
$2\times 3=6$ ......(ii)
Now, among these multiples of 3, the number of multiples of 3 which contain only the third power of 3, i.e ${{3}^{3}}=27$ are:
$27$
Thus, there is only one multiple of 3 which contain only the third power of 3.
So, the contribution for the exponent of 3 in N from this will be:
$3\times 1=3$ .....(iii)
From (i), (ii), and (iii), the total exponent of 3 in N will be the sum of all the contributions.
Thus, the exponent of 3 in N is:
$\begin{align}
  & 8+6+3 \\
 & \Rightarrow 17 \\
\end{align}$
Thus, the exponent of 3 in N is 17.
Now, we will in the same way find the exponent of 5 in N.
Now, among all the numbers which are being multiplied to form N, the multiples of 5 are:
$35,30,25,20,15,10,5,20$ (20 is repeated because 36! Is being multiplied by 20 to form N)
Now, among these multiples of 5, the multiples of 5 which contain only first power of 5, i.e. which are not multiples of 25 are:
$35,30,20,15,10,5,20$
Thus, there are a total of 7 multiples of 5 which contain only the first power of 5.
So, the contribution of the exponent of 5 in N from these numbers will be 7. ......(iv)
Now, among these multiples of 5, the multiples containing only the second power of 5, i.e. the multiples of 25 are:
$25$
Thus, there is only one multiple 5 which contains the second power of 5.
So, the contribution of the exponent of 5 in N from these numbers will be:
$2\times 1=2$ .......(v)
From (iv) and (v), the total exponent of 5 in N is the sum of all the contributions of the exponent of 5.
Thus, the exponent of 5 in N is:
$\begin{align}
  & 7+2 \\
 & \Rightarrow 9 \\
\end{align}$
Thus, the exponent of 5 in N is 9.
Now, we have obtained both the exponents of 3 and 5 in N.
We will now subtract them and get our answer.
Thus, the difference between the exponents of 3 and the exponent of 5 in N is given as:
$\begin{align}
  & 17-9 \\
 & \Rightarrow 8 \\
\end{align}$
Thus, 8 is our required difference.

Note: Be careful to count the different powers of 3 and 5. This is a very important step which cannot be missed otherwise the solution will be completely wrong. We can take all the multiples together and also and count them according to their powers in 3. For example, when we took all the multiples of 3 present, we could have counted 9,18 and 36 twice as they have two powers of 3 as factors and 27 three times and the rest one time.