Question 6 of 12

Question 6 (10 marks)

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. The 12-letter word TORONTONIANS with letter multiplicities T:2, O:3, R:1, N:3, I:1, A:1, S:1.

Find. (a) Total distinct permutations. (b) Count starting with T and ending with S. (c) Count starting with a vowel. (d) Count containing the fixed sub-block NOTNOT in order. (e) Count with the same letter first and last.

Approach. Use the multiset-permutation formula $\dfrac{n!}{\prod k_i!}$ throughout, fixing letters/blocks as required and re-permuting what remains.

1. (a) Total permutations. $n=12$ letters with repeats T(2), O(3), N(3), and R,I,A,S each appearing once: $$\dfrac{12!}{2!\,3!\,3!} = \dfrac{479{,}001{,}600}{2\cdot 6\cdot 6} = \dfrac{479{,}001{,}600}{72}$$ $\boxed{6{,}652{,}800}$
2. (b) Start with T, end with S. Fix one T at the front and the (unique) S at the back; the remaining 10 letters — T(1), O(3), N(3), R, I, A — fill the middle in any order: $$\dfrac{10!}{1!\,3!\,3!} = \dfrac{3{,}628{,}800}{36}$$ $\boxed{100{,}800}$
3. (c) Start with a vowel. Vowels present: O (×3), I (×1), A (×1) — 5 vowel "slots" out of 12 letter positions total. By the symmetry of multiset permutations, each of the 12 letter-occurrences is equally likely to occupy the first position across all $6{,}652{,}800$ arrangements, so the fraction starting with a vowel is $5/12$ (letters of the same kind are indistinguishable, so grouping by type gives an exact proportional count, not merely a probability estimate): $$6{,}652{,}800\times\dfrac{5}{12} = 554{,}400\times 5$$ $\boxed{2{,}772{,}000}$
4. (d) Contains the block NOTNOT (fixed order). The block NOTNOT consumes N(2), O(2), T(2) as one indivisible unit. What remains from the full multiset T(2),O(3),N(3),R,I,A,S after removing N(2),O(2),T(2) is: O(1), N(1), R, I, A, S — six letters, each appearing exactly once. Treating the block as one "super-letter", there are $1+6=7$ distinct items (all with multiplicity 1) to permute freely: $$7! = 5040$$ $\boxed{5040}$
5. (e) Same letter first and last. Only letters with multiplicity $\ge 2$ can occupy both end positions: T(2), O(3), N(3). For each such letter $L$, use two of its copies at the ends and permute the remaining 10 letters (with $L$'s count reduced by 2): $$T:\ \dfrac{10!}{3!\,3!}=100{,}800\qquad O:\ \dfrac{10!}{2!\,3!}=302{,}400\qquad N:\ \dfrac{10!}{2!\,3!}=302{,}400$$ These three cases are mutually exclusive (the end-letter is fixed per case), so they add: $$100{,}800+302{,}400+302{,}400$$ $\boxed{705{,}600}$