Q: (a) In how many ways can 7 men and 7 women be seated around a circular table so that no two men/no two women sit next to each other ?

(b) Suppose that at a sports dinner we have 16 cricketers and 6 tennis players. In how many ways can we seat them at a long table if (i) none of the tennis players is seated next to another tennis player and (ii) all tennis players are seated together.

(c) There are 20 persons among whom two are brothers. Find the number of ways in which we can arrange them around a circle so that there is exactly one person between the two brothers.

(d) A tea party is arranged for 2m people along two sides of a long table with m chairs on each side. r men wish to sit on one particular side and s on the other. In how many ways can they be seated? (Assume that r, s ≤ m.)

(e) A family consists of a grandfather, m sons and daughters and 2n grandchildren. They are to be seated in a row for dinner. The grandchildren wish to occupy the n seats at each end and the grandfather refuses to have a grandchild on either side of him. In how many ways can the family be made to sit?

**Solution:** (a) First arrange the 7 men in 6! ways. Then there are 7 places in between the men which will be occupied by 7 women in 7!, ways. So total no. of ways = 6! × 7!

(b) (i) First, 16 cricketers can be seated in 16! ways. There are 17 places in between them out of which 6 have to be occupied by tennis players.

No. of ways = ^{17}P_{6}

Required no. of ways = 16! × ^{17}P_{6}

(ii) All tennis players can be treated as one set. So now we have a total of 17 players. Their no. of permutations = 17 !

Also 6 tennis players can be arranged in 6! ways.

No. of ways = 17 ! × 6!

(c) We can arrange 18 persons around a circle in (18 –1)! = 17! ways. Now, there are exactly 18 places where we can arrange the two brothers. Also the two brothers can be arranged in 2! ways.

Thus, the number of ways of arranging the persons subject to the given condition is (17!) (18) (2!) = 2(18!).

(d) We can arrange r persons on m chairs on a particular side in ^{m}P_{r} ways and s persons on m chairs on the other side in ^{m}P_{s} ways. We can arrange (2m –r –s ) persons on the remaining (2m –r –s ) chairs in ^{2m-r-s}P_{2m-r-s} ways.

Thus, the number of ways of arranging the persons subject to the given conditions is

2(^{m}P_{r}) (^{m}P_{s}) (^{2m-r-s}P_{2m-r-s} ).

(e) The total number of seats required at the table is 1 + m + 2n . The grand children can occupy the n seats on either side of the table in ( ^{2n}P_{2n} ) ways. The grandfather can occupy a seat in ^{m-1}P_{1} ways. The remaining seats can be occupied in ^{m}P_{m} ways.

Therefore, the required numbers of ways is

= ( ^{2n}P_{2n} ) ( ^{m}P_{m} ) (^{m-1}P_{1})

= (2n!) m! (m –1)