Six boxes are numbered 1, 2, 3, 4, 5 and 6. Each box must contain either a white ball or a black ball. At least one box must contain a black ball and boxes containing black balls must be consecutively numbered. find the total number of ways of placing the balls.
If there is 1 black ball, it can be placed in 6 ways.
If there are 2 black balls, they can be placed in 5 ways (in 1,2 ; 2,3 ; 3,4 ; 4,5 and 5,6) and so on.
If there are 6 black balls, they can be placed in 1 way.
The total number of ways of placing the balls is
= 1 + 2 + 3 + 4 + 5 + 6
In how many ways can 6 green toys and 6 red toys be arranged, such that 2 particular red toys are never together whereas 2 particular green toys are always together?
Considering two green toys that 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!
There are five comics numbered from 1 to 5. In how many ways can they be arranged, so that part-1 and part-3 are never together?
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!
Therefore, the total number of arrangements, in which they are not together is:
= 120 - 48
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:
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 arranged in a particular game in 2 ways.
Total number of arrangements
= 36 × 21 × 2
= 1512 ways
Ten coins are tossed simultaneously. In how many of the outcomes will the third coin turn up a head?
When a coin is tossed once, there are two outcomes. It can turn up a head or a tail.
When 10 coins are tossed simultaneously, the total number of outcomes = 210
Out of these, if the third coin has to turn up a head, then the number of possibilities for the third coin is only 1 as the outcome is fixed as head.
Therefore, the remaining 9 coins can turn up either a head or a tail = 29