Answer: B let assume that there are 64 identical balls which are to be arranged in 5 different compartments (Since a,b,c,d,e are distinguishable) If the balls are arranged in a row i.e, o,o,o,o,o,o......(64 balls) We have 63 gaps where we can place a wall in each gap, since we need 5 compartments we need to place only 4 walls. We can do this in 63C4 ways.
Q. No. 14:
Six boxes are numbered 1,2,34,5 and 6. Each box must contain either a white ball or a black ball. Atleast one box must contain a black ball and boxes containing black balls must be consecutively numbered. find the total number of ways of placing the balls.
Answer: C If there is 1 black ball, it can be placed in 6 ways. If there are 2 black balls, they can be placed in 5 ways (in 1,2 ; 2,3 ; 3,4 ; 4,5 and 5,6) and so on. If there are 6 black balls, they can be placed in 1 way. The total number of ways of placing the balls is 1+2+3+4+5+6 =21.
Q. No. 15:
I have an amount of Rs 10 lakh, which I went to invest in stocks of some companies. I always invest only amounts that are multiples of Rs 1 lakh in the stock of any company. If I can choose from among the stocks of five different companies, In how many ways can I invest the entire amount that I have?
Answer: C The situation is similar to placing 10 identical balls among 5 distinguishable boxes, where a box may have zero or more balls in it. This case can be represented as arranging ten balls and (5-1) four walls in the single row, which can be done in 14C4 ways.(The balls placed between every successive pair of walls belong to one group) 14C4 = 1001 ways.
Q. No. 16:
In a cricket match if a batsman score 0,1,2,34,or 6 runs of a ball, then find the number or different sequences in which he can score exactly 30 runs of an over. Assume that an over consists of only 6 balls and there were no extra and no run outs.
Answer: B Case A:- Five 6 and one 'zero' = 6!/5! = 6. Case B:- Four 6 and one '2' and one '4' : 6!/4! = 30. Case C:- Three 6 and three '4' = 6!/(3!*3!)= 20 Case D:-Four 6 and two '3' = 6!/(4!*2!) = 15. Total number of different sequences = 71.
Q. No. 17:
A book-shelf can accommodate 6 books from left to right. If 10 identical books on each of the languages A,B,C and D are available, In how many ways can the book shelf be filled such that book on the same languages are not put adjacently.
Answer: D First place can be filled in 4 ways. The subsequent places can be filled in 3 ways each. Hence, the number of ways = 4*3*3*3*3*3 = 4 * 35
Q. No. 18:
What is the total number of ways in which Dishu can distribute 9 distinct gifts among his 8 distinct girl friends such that each of them gets at least one gift?
Answer: C As every girl friend should get one gift. The number of ways 8 distinct gifts can be selected is 9C8 = 9ways. The number of ways each GF gets one gift each out of these 8 selected gifts 8! Total number of ways 8 gifts can be distributed is 9 * 8!. Now the last one gift can be given to any of the 8 GF hence total number of ways of distributing = (9 * 8! * 8)/2 = 36 * 8! ways.