Question 4 of 12

Question 4

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A pool of 12 men and 12 women (24 people total), forming committees of size 8; and the 11-letter word ABBACADABRA with letter multiset $\{A^5,B^3,C,D,R\}$.

Find. (a) the 3 constrained committee counts; (b) the number of distinct rearrangements, unrestricted and with all A's together.

Approach. Committees: use complementary counting for "at least one of each" and "not both together"; direct product for "equal numbers". Word arrangements: multinomial coefficient for the unrestricted count; glue the 5 A's into one block to force togetherness.

1. 4a-i) At least one man and one woman. Total 8-person committees from 24 people minus the two excluded extremes (all-men, all-women): $$\binom{24}{8}-\binom{12}{8}-\binom{12}{8}=735{,}471-495-495=734{,}481$$ $\boxed{734{,}481}$ committees.
2. 4a-ii) Equal numbers of men and women. Exactly 4 men and 4 women: $$\binom{12}{4}\binom{12}{4}=495\times495=245{,}025$$ $\boxed{245{,}025}$ committees.
3. 4a-iii) John and Rob never together. Subtract committees containing *both* John and Rob (choose the remaining 6 members from the other 22 people) from the unrestricted total $\binom{24}{8}$: $$\binom{24}{8}-\binom{22}{6}=735{,}471-74{,}613=660{,}858$$ $\boxed{660{,}858}$ committees.
4. 4b-i) ABBACADABRA, no restrictions. The word has 11 letters with multiplicities $A{:}5,\,B{:}3,\,C{:}1,\,D{:}1,\,R{:}1$; the multinomial (permutations-of-a-multiset) count is $$\frac{11!}{5!\,3!\,1!\,1!\,1!}=\frac{39{,}916{,}800}{120\cdot6}=55{,}440$$ $\boxed{55{,}440}$ arrangements.
5. 4b-ii) All A's together. Glue the 5 A's into one block; the sequence to arrange is now the block plus $B,B,B,C,D,R$ — 7 items with 3 identical B's: $$\frac{7!}{3!}=\frac{5{,}040}{6}=840$$ $\boxed{840}$ arrangements.