Question

All possible numbers are formed using the digits 5, 5, 4, 4, 4, 4, 4, 5, 6, 6, 6 taken all at a time. Let the number of such numbers in which odd digits will occupy even places is $n$. Then the sum of digits of $n$ is

Answer 1

11

Answer 2

We have 11 digits in total: three odd digits (all 5’s) and eight even digits (five 4’s and three 6’s).

1. Arrange Odd Digits:
   The even positions in an 11-digit number are: 2, 4, 6, 8, 10 (5 positions). The condition is that every odd digit must go into an even position.

   • Choose 3 out of 5 positions for the odd digits: $$\binom{5}{3} = 10.$$
   • Since all odd digits are 5, they can be arranged in only 1 way.

2. Arrange Even Digits:
   The remaining positions (the 6 odd positions and the 2 even positions left) get the eight even digits.

   • The even digits include 5 copies of 4 and 3 copies of 6.
   • Arrangements for these digits: $$\frac{8!}{5!\,3!} = \frac{40320}{120 \times 6} = 56.$$

3. Total Arrangements (n):
   Multiply the number of ways:

   $$n = \binom{5}{3} \times 1 \times \frac{8!}{5!3!} = 10 \times 56 = 560.$$

4. Sum of the digits of n:
   For $560$, the sum is $5+6+0=11$.