Question

A four-digit number of distinct digits is formed by using the digits 2, 3, 4, 5, 6, 7, 8. The number of such number which are divisible by 25 is
A.60
B.40
C.20
D.15

Answer 1

Hint: In this question, we will first form a number from the given digits which is divisible by 25; for example, 25 and 75. We will now put these numbers at the last places of four-digits. Then we would have only 5 digits to place in the rest of the 2 places. So, we can write it as $ ^5{P_2} $ . Let’s see how we can solve it.

Complete step-by-step answer:
The numbers which are divisible by 25 if their last two digits are divisible by 25.
We have 2, 3, 4, 5, 6, 7, 8 in which we can form 25 and 75 which are divisible by 25.
We will put 25 at least two places of four-digits number.
For remaining two places out of 5-digits can be filled in
  $ \Rightarrow ^5{P_2} = \dfrac{{5!}}{{(5 - 2)!}} = \dfrac{{5!}}{{3!}} $ = 20 ways
Similarly, we will put 75 at least two places of four-digits number.
For remaining two places out of 5-digits can be filled in
 $ \Rightarrow ^5{P_2} $ = 20 ways
Thus, a total number of four-digits can be formed = 20 + 20 = 40 (Ans.)

So, the correct answer is “Option B”.

Note: There is one thing you need to keep in mind, here we can form only two numbers that is 25 and 75 from the given digits, that is why we have calculated 20 ways when we put 25 at the last two-places and Similarly we have calculated 20 ways for 75. But suppose if we have 0 in the digits then we can form 50 also, because 50 is also divisible by 25. So, again we need to calculate by using permutation. So, our new answer would be 20 + 20 + 20 = 60.
We know that $ ^n{P_r} = \dfrac{{n!}}{{(n - r)!}} $