Some boys are standing on a circle at distinct points. Each possible pair of persons, who are not adjacent, sing a 3 minute song, one pair after another. The total time taken by all the pairs to sing is 1 hour. Find the number of boys?
Answer: C Each boy would pair with n-3 other boys Number of possible pairs = n(n-3)/2 n(n-3)/2 = 60/3 = 20 => n(n-3)= 40 => n = 8
Q. No. 68:
There are 10 seats around a circular table. If 8 men and 2 women have to seated around a circular table, such that no two women have to be separated by at least one man. If P and Q denote the respective number of ways of seating these people around a table when seats are numbered and unnumbered, then P/Q equals
Answer: C Initially we look at the general case of the seats not numbered. The total number of cases of arranging 8 men and 2 women, so that women are together => 8!2! The number of cases where in the women are not together => 9! - (8!2!) = Q Now, when the seats are numbered, it cab be considered to a linear arrangement and the number of ways of arranging the group such that no two women are together is => 10! - (9!2!). But the arrangements where in the women occupy the first and the tenth chairs are not favourable as when the chairs which are assumed to be arranged in a row are arranged in a circle, the two women would be sitting next to each other. The number of ways the women can occupy the first and the tenth position = 8!2! The value of P = 10! - (9!2!) - (8!2!) Thus P/Q = 10/1
Q. No. 69:
In how many ways can the letters ABACUS be rearranged such that vowels always appear together ?
Answer: D The three vowels in ABACUS are A,A and U. These three can be arranged among themselves in 3!/2! = 3 ways. As the three vowels are to be appear together, we consider them as one entity. Thus we have four letters: (AAU), B,C and S to be arranged. This can be arranged in 4! ways. Required number of ways = (4!*3!)/2.
Q. No. 70:
A five-digit number divisible by 3 is to be formed using numerical 0,1,2,3,4 and 5 without repetition. The total number of ways this can be done is :
Answer: C Using the digits 0,1,2,3,4 and 5, five digits numbers divisible by 3, can be formed using the following combinations. Case I : 1,2,3,4,5 Total number of numbers formed using three digits = 5! = 120. Case II : 0,1,2,4,5 Total number of numbers formed using three digits = 4*4*3*2 = 96. Thus, total number = 120+96 = 216.
Q. No. 71:
Amit has 11 friends : 7 boys and 4 girls. In how many ways, can Amit invite them, if there have to exactly 4 boys in the invitees ?
Answer: A Case I : Number of ways of inviting friends, when no girls invited =7C4*4C0 = 35 ways. Case II : Number of ways of inviting friends when 1 girl is invited = 7C4*4C1 = 140 ways. Case III : Number of ways of inviting the friends, when two girls are invited = 7C4*4C2 = 210 ways. Case IV : Number of ways of inviting the friends, when three girls are invited = 7C4*4C3 = 140 ways. Case V : Number of ways of inviting the friends, when four girls are invited = 7C4*4C3 = 35 ways. So, Total number of ways = 35+140+210+140+35 = 560 ways.
Out of the 6 ruling and 5 opposition party members, 4 are to be selected for a delegation.
Q. No. 1:
In how many ways can this be done so as to include at least one opposition member ?
Answer: B Selecting one opposition party member =5C1*6C3=5*20 = 100 ways Selecting two opposition party members = 5C2*6C2=10*15 = 150 ways Selecting three opposition party = 5C3*6C1 = 10*6 = 60 ways Selecting all four opposition party members = 5C4 = 5 ways. Total ways = 100+150+60+5 = 315 ways.
Q. No. 2:
In how many ways can this be done so as to include exactly one ruling party member?
Answer: D Number of ways of selecting exactly one ruling party member =6C1 = 6 ways. Number of ways of selecting remaining three members from opposition party = 5C3 = 10 ways. Total ways = 6*10 = 60 ways.