سوال ۱۱۳ کنکور ارشد علوم کامپیوتر ۱۴۰۴

The word $AAABBCCCDD$ is a multiset with $3$ A's, $2$ B's, $3$ C's and $2$ D's, for $10$ letters in total.

Total arrangements. Counting all distinguishable permutations of the multiset: $\frac{10!}{3!\,2!\,3!\,2!}=\frac{3628800}{6\cdot2\cdot6\cdot2}=\frac{3628800}{144}=25200.$

Subtract those containing the substring $AB$. Use inclusion–exclusion on the number of glued $AB$ blocks. Forming $j$ copies of the super-letter "$AB$" uses up $j$ A's and $j$ B's, leaving $3-j$ A's and $2-j$ B's. Since we have only $2$ B's, at most $j=2$ blocks are possible. The number of arrangements that contain (at least) $j$ chosen disjoint $AB$ blocks is $S_j=\frac{(10-j)!}{(3-j)!\,(2-j)!\,3!\,2!\,j!},$ where the extra $j!$ accounts for the $j$ identical $AB$ blocks.

$S_1=\frac{9!}{2!\,1!\,3!\,2!\,1!}=\frac{362880}{2\cdot1\cdot6\cdot2}=15120,$ $S_2=\frac{8!}{1!\,0!\,3!\,2!\,2!}=\frac{40320}{1\cdot1\cdot6\cdot2\cdot2}=1680.$

Arrangements with no $AB$ substring (inclusion–exclusion $S_0-S_1+S_2$): $t=25200-15120+1680=11760.$

Final value. $\frac{t}{240}=\frac{11760}{240}=49.$

Since $30<49\le 50$, the value lies in $(30,50]$ — option 3.

Exam trick: For a forbidden adjacency in a multiset, glue the pattern into a super-letter and apply inclusion–exclusion over the number of glued copies; remember to divide by $j!$ because the identical blocks are interchangeable.