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 Case A:- If '0' is at the unit's place, then the number of ways of three-digit integer = 3*4 =12 ways. Case B:- If '5' is at the unit's place and '0' is at the ten's place, then the number of ways of three digit integers = 3 ways. Case C:- If '5' is at unit's place and '0' is not at the ten's place, then the number of ways of three-digit integers =2*3 = 6 ways. So, total number of ways = 12+3+6 = 21 ways.
Q. No. 20:
There are five intermediate hurdles between two points A and B. In how many ways can a man stop at three of these hurdles such that no two halts are at consecutive hurdles?
Answer: D Let the three hurdles where the man stops be denoted by H1, H2 and H3and let us denote the number of Hurdles before H1 by a0, the number of hurdles between H1 and H2 by a1, the number of hurdle between H2 and H3 by a2 and the number of hurdle after H3 by a3 respectively. a0 +a1 + a2 + a3 =12 where, a0, a3 >=0 and a1, a2 >=1. Let a0+1 =A0, a3 + 1 =A3 Thus, A0 +a1+a2+a3+A3 = 14. Number of possible solutions to the above equation is (14-1)C3 = 286 ways. The number of ways in which the man can stop at three hurdle such that no two of them are consecutive is 286.
Q. No. 21:
There are 20 couples in a party. Every person greets every person except his or her spouse. People of the same sex shake hands and those of opposite sex greet each other with a namakar (It means bringing one's own palms together and raising them to the chest level). What is the total number of handshakes and namaskar's in the party?
Answer: B There are 20 men and 20 women. When a man meets a women, there are two namaskars, whereas when a man meets a man (or a women) there is only 1 handshake. Number of handshakes = 2(20C2) (men and women ) = 380. For number of Namskars Every man does 19 namaskars (to the 20 women excluding his wife) and they respond in the same way. Number of Namaskars = 2(20)(19) = 760 Total =760+380=1140
Q. No. 22:
How many 6 lettered words can be formed using the letters P,Q,R,S,T and U such that all the words begin with P and no two letters of P, S and U are adjacent. Repetition of letters is not allowed.
Answer: C There is only 1 choice of for the first letter, namely P. Also, the second letter cannot be S and U. So, there are 3 choices for it. In general, the remaining 4 letters could be arranged in 4! ways. However, we must discount those cases wherein S and U are adjacent. Considering S and U as one object, then, we subtract 3!2!words (as D and F can be arranged among themselves in 2! ways). Total number of words = 3*(4!-3!2!) = 36.
Q. No. 23:
In a GPLassets corporation, there are 20 technicians and 5 marketing managers. How many committees can be formed of 6 technicians and 3 marketing managers?
Answer: A The technicians can be selected in 30C6 ways and the marketing managers in 5C3 ways. Total number of ways = 30C6 * 5C3 = 387600 ways.
Q. No. 24:
A student is required to answer 6 out of 10 questions divided into two groups each containing 5 questions. He is not permitted to attempt more than 4 from each group. In how many ways can he make the choice?
Answer: D Number of ways of choosing 6 from 10 = 10C6 =210 Number of ways of attempting more than 4 from a group =2*5C5*5C1= 10 Required number of ways = 210 - 10 =200