A committee is to be formed comprising 7 members such that there is a simple majority of men and at least 1 women. The shortlist consists of 9 men and 6 women. In how many ways can this committee be formed?

Answer: C Three possibilities:- 1W+6M, 2W+5M, 3W+4M =>(6_{C1}*9_{C6}) +(6_{C2}*9_{C5}) + (6_{C3}*9_{C4}) = 4914

Q. No. 2:

For a GPLassets, a selection committee is to be chosen consisting of 5 ex-technicians. Now there are 12 representatives from four zones. It has further been decided that if Dishant is selected, Gautam and Jitendra will not be selected and vice-versa. In how many ways it can be done?

Answer: D In a leap year there are 366 days i.e 52 weeks + 2 extra days. So to have 53 Sundays one of these two days must be a Sunday. This can occur in only 2 ways. i.e (Saturday and Sunday) or (Sunday and Monday). Thus number of ways =2.

Q. No. 4:

A tea expert claims that he can easily find out whether milk or tea leaves were added first to water just by tasting the cup of tea. In order to check this claims 10 cups of tea are prepared, 5 in one way and 5 in other. Find the different possible ways of presenting these 10 cups to the expert.

Answer: A Since there are 5 cups of each kind, prepared with milk or tea leaves added first, are identical hence, total number of different people ways of presenting the cups to the expert is10!/(5!*5!) = 252.

Q. No. 5:

A 6x6 grid is cut from an 8x8 chessboard. In how many ways can we put two identical coins, one on the black square and one on a white square on the grid, such that they are not placed in the same row or in the same column?

Answer: A In a 6x6 grid of a chessboard, each row and each column contains 3 white and 3 black squares placed alternatively. There are a total of 18 black and 18 white squares. For every black square chosen to put one coin, we cannot choose any white square present in its row or column. There are 3 white squares in its row and 3 white square in its column for every black square. Hence for every black square chosen, we can choose (18-6) =12 white squares. Total number of possibilities where a black square and a white square can be chosen so that they do not fall in the same ro or in the same column = 18*12 = 216. So there are 216 ways of placing the coins that are identical.

Q. No. 6:

There are eight boxes of chocolates, each box containing distinct number of chocolates from 1 to 8. In how many ways four of these boxes can be given to four persons (one boxes to each) such that the first person gets more chocolates than each of the three, the second person gets more chocolates than the third as well as the fourth persons and the third person gets more chocolates than fourth person?

Answer: B All the boxes contains a distinct number of chocolates. For each combinations of 4 out of 8 boxes, the box with the greatest number has to be given to the first person, the box with the second highest to the second person and so on. The number of ways of giving 4 boxes to the 4 person is 8_{C4} = 70