Permutation of Digits & Divisibility

This is an interesting counting methods question that combines a very basic rule of test of divisibility of number by 3.

Question

How many six digit positive integers comprising only the digits 1 and 2 can be formed such that the number is divisible by 3?

Correct Answer

Choice C. 22 such numbers can be formed.

Video Explanation

Explanatory Answer

Test of divisibility by 3

If a number is divisible by 3, the sum of the digits of the number must be divisible by 3.

In this question we are forming 6 digit numbers comprising only two digits: 1 and 2.

We will list down various possibilities.

Case 1: 111111: All digits are 1s. The sum of the digits of the number is 6, which is divisible by 3. This number can rearrange in only one way.

Case 2: 222222: All digits of this number are 2s. The sum of the digits is 12, which is divisible by 3. This number also can rearrange in only one way.

Case 3: 111222: In case 1, where all digits are 1, the number was divisible by 3. If we remove 3 ones from 111111 and replace it with 3 twos, the resultant number will be divisible by 3.

This number can rearrange in $\frac{6!}{3!\times 3!}$ = $\frac{720}{36}$ = 20 ways