Porisma

Worked solution

Keep the meeting and lunch seating models separate

(a) Find the number of ways the letters of POSITIVE can be arranged if

(i) all vowels must be next to each other, marks

(ii) the vowels are separated and the arrangement begins and ends with I. marks

(b) The Mathematics Department of M Institute consists of 11 teachers, including Mr Tan and Mr Lin. At a meeting, all 11 teachers sit at a round table with numbered seats. How many arrangements have Mr Tan and Mr Lin next to each other? marks

After the meeting they find a circular table for 6 and a long table with a row of 5 seats at a food court, as shown. Find the number of arrangements of all 11 teachers at these two tables such that Mr Tan is seated next to Mr Lin. marks

[2011/MI/II/8]

CJC · Review problems · Question 11

Answer

(a)(i) 1440; (ii) 144. Meeting: 7983360; lunch: 1209600.

(a)(i) Group the vowels

The vowels are O,I,I,E and the four consonants are P,S,T,V. Arrange the vowels in 4!/2!4!/2! ways and their block alongside four consonants in 5!5! ways: (4!/2!)×5!=1440(4!/2!)\times5!=1440.

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(a)(ii) Fix I at each endpoint

Arrange the four consonants in 4!4! ways between the endpoint I’s. The remaining vowels O and E must go in two of the three internal consonant gaps: (32) 2!\binom32\,2! ways. The two gaps next to an endpoint I are forbidden. Total 4!(32) 2!=1444!\binom32\,2!=144.

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(b) Meeting: numbered circular seats

The Tan–Lin block plus 9 other teachers makes 10 circular objects. Their arrangements are (10−1)!(10-1)!, with 2!2! internal pair orders. Numbered seats give 11 placements of a reference person: (10−1)! 2!×11=7983360(10-1)!\,2!\times11=7983360.

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Lunch case 1: pair at the circular six-seater

Choose four companions from the other 9 teachers, then arrange the pair block plus those four in a circle: (94)(5−1)!×2\binom94(5-1)!\times2. Arrange the other 5 in a row in 5!5! ways. This gives 725760.

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Lunch case 2: pair at the linear five-seater

Choose three companions from 9. The pair block plus three companions has 4!4! row orders and 2 internal orders. Arrange the remaining six circularly in (6−1)!(6-1)! ways: (93) 4!×2×(6−1)!=483840\binom93\,4!\times2\times(6-1)!=483840. The disjoint total is 725760+483840=1209600725760+483840=1209600. The lunch circle is not specified as numbered, so the meeting’s 11-seat multiplier does not carry over.

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Six seats around a circular table are separate from a single straight row of five seats at a long table, retaining the source layout and capacities. The lunch circle is unnumbered.
Six seats around a circular table are separate from a single straight row of five seats at a long table, retaining the source layout and capacities. The lunch circle is unnumbered.
Six seats around a circular table are separate from a single straight row of five seats at a long table, retaining the source layout and capacities. The lunch circle is unnumbered.

Common mistakes

The meeting uses numbered seats; lunch does not. The tables also have different capacities.

Solution notes

All four printed allocations, complete meeting/lunch working, both lunch cases and source table layout retained. No condition of the meeting’s labelled seats is silently transferred to lunch.

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