3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 3 · AdvancedSingle correct
Letters are taken from the word EXAMINATION, each letter being used at most as many times as it occurs in the word, and arranged in a row to form a 4-letter string (strings need not be meaningful; strings that look identical are counted once). The number of distinct such strings is ______. Exactly one option is correct.
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Answer: B
EXAMINATION has A, I, N twice each and E, X, M, T, O once: 8 distinct letters, of which 3 are available twice and none three times. Split by the repetition pattern of the string (cases are exhaustive and disjoint).
(i) All four letters different: 8P4=1680.
(ii) Exactly one letter used twice: choose it (3 ways), choose the two other letters from the remaining 7 ((27)=21), arrange: 2!4!=12. Total 756.
(iii) Two letters used twice: (23)=3, arrangements 2!2!4!=6. Total 18.
Answer 1680+756+18=2454.
Trap: in case (ii), choosing the other two letters from all 8 letters ((28)=28, giving 1008) is wrong, because one of them would be the doubled letter itself (a triple) and the result 2706 is an overcount; omitting case (iii) gives 2436.
Difficulty: Very hardQuestion 2Level 3 · AdvancedSingle correct
Five boys B1,…,B5 and five girls G1,…,G5 sit around a round table with 10 identical seats; two seatings are the same if one is a rotation of the other (reflections are different). Boys and girls must alternate, and G1 must not sit next to B1. The number of seatings is ______. Exactly one option is correct.
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Answer: C
Fix B1's seat; this removes the rotational symmetry, after which seats are distinguishable. The other boys take the 4 remaining alternate seats in 4! ways. The 5 girls' seats are the seats between consecutive boys; exactly 2 of them are adjacent to B1.
G1 has 3 allowed seats, and the other girls fill the rest in 4! ways.
Total =4!⋅3⋅4!=1728. (Check: all alternating seatings 4!5!=2880; those with G1 next to B1: 4!⋅2⋅4!=1152; 2880−1152=1728.)
Difficulty: Very hardQuestion 3Level 3 · AdvancedSingle correct
In how many distinct arrangements of all eleven letters of the word MISSISSIPPI are no two S's adjacent and no two I's adjacent?
Exactly one option is correct.
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Answer: A
The gap method cannot be used twice independently (placing the I's into gaps can create S's that are adjacent once other letters are removed, and vice versa). Instead first write the word formed by the four I's and four S's alone, then insert M, P, P.
An I/S word with four of each and r runs has e=8−r adjacent equal pairs, and each such pair must receive at least one of M, P, P. Since only 3 letters are available, e≤3, i.e. r≥5. Counting by runs (compositions of 4): r=8: 2 words; r=7: 2(33)(23)=6; r=6: 2(23)2=18; r=5: 2(23)(13)=18. So e=0,1,2,3 occur 2,6,18,18 times.
Insertion: there are 9 gaps (including ends); if gj letters go into gap j, ∑gj=3 with the e bad gaps ≥1: (83−e+8)=(811−e) ways, i.e. 165,45,9,1. The inserted letters can be in 2!3!=3 orders.
Total =3(2⋅165+6⋅45+18⋅9+18⋅1)=3⋅780=2340.
(7350 counts only the S-condition.)
Difficulty: Peak challenge