In a class of 10 students, there are 3 girls. In how many different ways can they be arranged in a row such that no two of three girls are consecutive?
Answer: D Since there is no restriction on the arrangement of boys. 7 boys can be arranged in a row in 7! ways. Now there are 8 places of 3 girls. Therefore, 3 girls can be arranged in 8P3 Ways. Hence required number of ways = 8P3 * 7!
Q. No. 32:
A new flag is to be designed with six vertical stripes using some or all of the colours yellow, green, blue and red. Then, the number of ways this can be done so that no two adjacent stripes have the same colour is
Answer: A The first strip can be of any of the four colours, The 2nd can be of any colour except that of the first (i.e. 3). Similarly, each subsequent strip can be of any colour except that of the preceding strip (=3) Hence number of ways = 4 × 35 = 12 × 81
Q. No. 33:
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. 34:
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. 35:
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.