Question

Consider ${}^{\prime }abcde{f}^{\prime }$ as a $6$ digit number. If $n$ is the number of such numbers whose sum of digits is $m$, then which of the following option(s) is/are correct?

• A. If $m=54$, then $n=6!$
• B. If $m=54$, then $n=1$
• C. If $m=53$, then $n=6$
• D. If $m=52$, then $n=21$

Answer 1

Answer: (B), (C), (D)

If $m=52$, then $n=21$

Given number : $abcdef$
Maximum sum of digits can be $=6\times 9=54$

Case $\left(1\right)m=54$
Each digit must be $9$,
Only one possible number, $n=1$

Case $\left(2\right)m=53$
Possible digit can be $9,9,9,9,9,8$
$\therefore$Total possible numbers
$n=\frac{6!}{5!}=6$

Case $\left(3\right)m=52$,
Possible digit can be $(9,9,9,9,9,7)\text{or}(9,9,9,9,8,8)$
$\therefore$Total possible numbers
$n=\frac{6!}{5!}+\frac{6!}{4!\cdot 2!}=6+15=21$