In how many ways can 6 green toys and 6 red toys be arranged, such that 2 particular red toys are never together where as 2 particular green toys are always together?
Answer: D Considering two green toys who are to be together as one unit. We can arrange the 6 green toys and the remaining 4 red toys (excluding the 2 who are not to be together) is => 9!*2!*10C2*2! = 18* 10! ways.
Q. No. 26:
A teacher of 6 students takes 2 of his students at a time to a zoo as often as he can, without taking the same pair of children together more than once. How many times does the teacher go to the zoo?
Answer: B The total number of ways in which 5 part can be arranged = 5! =120. The total number of ways in which part-1 and part-3 are always together = 4!*2! = 48. Therefore, the total number of arrangements, in which they are not together is =>120-48 = 72.
Q. No. 29:
The number of ways which a mixed double tennis game can be arranged amongst 9 married couples if no husband and wife play in the same is:
Answer: B As per the question there are 9 married couples and no husband and wife should play in the same game: We know that in a mixed double match there are two males and two females. Step I: Two male members can be selected in 9C2= 36 ways Step II: Having selected two male members, 2 female members can be selected in 7C2 = 21 ways. Step III: Two male and two female members can arrangedin a particular game in 2 ways. Total number of arrangements = 36*21*2 = 1512 ways.
Q. No. 30:
While packing for a business trip Mr. Debashis has packed 3 pairs of shoes, 4 pants, 3 half-pants,6 shirts, 3 sweater and 2 jackets. The outfit is defined as consisting of a pair of shoes, a choice of “lower wear” (either a pant or a half-pant), a choice of “upper wear” (it could be a shirt or a sweater or both) and finally he may or may not choose to wear a jacket. How many different outfits are possible?
Answer: D Number of ways a pair of shoes can be selected = 3C1 = 3 ways. Number of ways "lower wear" can be selected = (3 + 4) = 7 ways Number of ways "upper wear" can be selected = 3 + 6 + 3 × 6 = 27 ways Number of ways jacket can be chosen or not chosen = 3 ways {No jacket, 1st jacket, 2nd jacket) => Total number of different outfits = 3 × 7 × 27 × 3 = 1701 ways.