سوال ۱۱۸ کنکور ارشد علوم کامپیوتر ۱۳۹۷

We form 6-digit numbers from the multiset of digits $\{4,4,4,1,1,1\}$. A number is fixed by choosing which $3$ of the $6$ positions hold the digit $4$ (the rest hold $1$), so the count of distinct numbers is

$\binom{6}{3}=20.$

None of these has a leading zero, so all $20$ are valid 6-digit numbers.

Per-position digit sum (symmetry argument). Fix one of the six positions. How many of the $20$ numbers carry a $4$ in that position? After placing a $4$ there, the remaining two $4$'s occupy $2$ of the other $5$ positions, giving $\binom{5}{2}=10$ ways. By the same count, a $1$ sits in that position in the other $10$ numbers.

So, summed over all $20$ numbers, the digit appearing in any single position contributes

$10\cdot 4 + 10\cdot 1 = 50.$

Add the place values. Each position's total of $50$ is weighted by its place value $10^{k}$:

$\text{Sum} = 50\,\bigl(10^{5}+10^{4}+10^{3}+10^{2}+10^{1}+10^{0}\bigr) = 50\times 111111 = 5555550.$

This matches option 1. (Cross-check: option 4, $360\times555555 = 199999800$, and option 3, $11111100$, are far too large.)

Exam trick: For "sum of all numbers from a multiset of digits," never list them. Find the per-position digit total by symmetry, then multiply by the repunit $\underbrace{11\ldots1}_{n}$ of the right length.