The total number of possible proper three-digit integers that can be formed using 0,1,3,4 and 5 without repetition such that they are divisible by 5 are :
Answer: B Numbers divisible by 5 will end with either '0' or '5'. Total numbers ending with '0' = 4*3*1 = 12 Total of numbers ending with '5' = 3*3*1 = 9 Total such numbers = 21.
Q. No. 74:
In a certain laboratory, chemicals are identified by a colour-coding system. There are 20 different chemicals. Each one is coded with either a single colour or a unique two-colour pair. If the order of colours in the pairs does not matter, what is the minimum number of different colours needed to code all 20 chemicals with either a single colour or a unique pair of colours ?
Answer: B Each one coded with either a single colour or unique two-colour pair. Therefore, total number of ways = n + nC2 Minimum number of different colour needed to code all 20 chemicals will be 6. => 6 +6C2 = 21.
On the shelf there are 3 books on Psychology, 3 books on Management and 4 books on Economics.
Q. No. 1:
In how many ways can the books be arranged, if the books on only Management are to be arranged together ?
Answer: D Suppose total management book be 1 unit, then there are 8 books, which are arrange in 8! ways. Also, 3 management books are arranged in 3! ways. Total number of ways = 3!*8! = 241920.
Q. No. 2:
In how many ways can all the books be arranged at random ?
Answer: C As per the given condition, number in the highest position should be either 6 or 7, which can be done in 2 ways. If the first digit is 6, the other digits can be arranged in 6!/2! = 360 ways. If the first digit is 7, the other digits can be arranged in 6!/(2!*2!) = 180 ways. Thus required possibilities for n = 360+180 = 540 ways.
Q. No. 77:
In how many ways can four letters of the word 'SERIES' be arranged ?
Answer: D The given word 'SERIES' contains 2S, 2E and rest are distinct. The number of ways of selecting the 4 letter and the number of arrangement are as follows : Case I : 4 letters are distinct = S,E,R,I = 4! = 24 ways. Case II : 2 letter are same and 2 letter are distinct = {SSRI, SSRE, SSIE, EERI, EERS and EEIS} = 4!/2! * 6 = 72 ways. Case III : Two are same of one kind and two are same of the other kind = SSEE = 4!/(2!*2!) = 6 ways. Total number of ways = 24 + 72 + 6 = 102.
Q. No. 78:
The number of ways in which a mixed double tennis game can be arranged amongst 9 married couples if no husband and wife play in the same game is :
Answer: B In a mixed double of tennis game there is a pair of man and woman on each ride. So, total 2 men and 2 women are in each game. There are 9 men and 9 women. Ways to select 2 men = 9C2 Wives of these 2 men cannot play in same game. So ways to select 2 women = 7C2 In a game these 4 persons can be paired by 2 types. Total number of ways = 9C2*7C2*2 = 1512.