2 boys and 3 girls are to be seated round a table with 5 seats. Each child occupies exactly one seat. In how many ways can this be done if(adsbygoogle = window.adsbygoogle || []).push({});

(a) the 2 boys must be seated together

(b) same as (a) but this time the seats are numbered

Solution

(a) ##\frac{4!}{4}2!##

(b) ##\frac{4!}{4}2!\times 5##

The "##\times 5##" in (b) seems to contradict the "##\div 4##" in (a).

Why ain't the answers

(a) ##4!2!\div 5##, (b) ##4!2!## or

(a) ##\frac{4!}{4}2!##, (b) ##4!2!##

Explanation for the "correct" solution

(a) After grouping the 2 boys into one group, we have 4 groups (with one girl in a group). The number of ways of arranging 4 items in a circle is ##\frac{4!}{4}## since ABCD, DABC, CDAB and BCDA are the same arrangement when placed in a circle. The 2 boys can be permutated within the "boy group". So we have ##\frac{4!}{4}2!##.

(b) If the seats are numbered, rotating an arrangement by one seat results in a new arrangement. We can do 5 such rotations. So we have ##\frac{4!}{4}2!\times 5##.

But if we ##\div 4## in (a) since we are rotating the 4 groups ABCD 4 times and considering these 4 arrangements to be the same, then in (b) shouldn't we ##\times 4## instead of ##5##?

On the other hand, if we first consider the seats to be numbered and hence ABCD, DABC, CDAB and BCDA are NOT the same arrangement, we have 4!2!. Then for (a), since the seats are not numbered and there are 5 rotations that produce the same arrangement when there are 5 seats, we have (a) ##4!2!\div 5##.

My issue is why does the correct answer consider the rotations of groups in (a) but not in (b)? In (b), it considers instead the rotations of seats (or persons).

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# I Permutations in a circle

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