Question

In how many ways can 3 prizes be distributed among 4 girls, when
(i) no girl gets more than 1 prizes ,
(ii) a girl may get any number of prizes,
(iii) no girl gets all the prizes?

Answer 1

Hint: In case one we have distributed prizes so that no girls get more than one prize. For prize 1 we have 4 girls. But for prize 2 we only have 3 girls because no girls should have more than 1 prize. Similarly, for prize 3 we have 2 girls. Total number of ways \[=4\times 3\times 2=24\] . In case second, for prize 1 we have 4 girls and we are given that a girl may get any number of prizes. So, we have 4 girls for prize 2 and prize 3 respectively. Total number of ways \[=4\times 4\times 4=64\] . For case 3, just deduct the ways in which all prizes that can be given to one girl are second. Total number of ways \[=64-4\] .

Complete step-by-step answer:
Refer to the diagram and solve it further.

i) Number of ways that prize 1 can be given to any 4 girls = 4 .
Number of ways that prize 2 can be given to any 3 girls = 3 .
Number of ways that prize 2 can be given to any 2 girls = 2 .
Total number of ways \[=4\times 3\times 2=24\] .
ii) Number of ways that prize 1 was given to girls = 4 .
Number of ways that prize 1 was given to girls = 4 .
Number of ways that prize 1 was given to girls = 4 .
Total number of ways \[=4\times 4\times 4=64\] .
iii) If no boys get all prizes then , we have to remove those ways from 64 in which a boy gets all prizes . Clearly, we can give all prizes to the first or second or third or fourth boy . We have 4 ways in which all the prizes are given to a boy.
So, after deduction we get\[=64-4=60\] .
Hence, the number of ways in which no girls get all the prizes is 60.

Note: Case one can also be solved in another way. We have to distribute 3 prizes to 4 girls. We need to select any 3 out 4 girls. The number of ways to select 3 girls is \[^{4}{{C}_{3}}\] . Also, rearrangement is possible.
So, the total number of ways \[{{=}^{4}}{{C}_{3}}\times 3!=4\times 3=12\] .