Question

There are three prizes to be distributed among $ 6 $ boys. In how many ways can it be done when
I.No boy gets more than one prize.
II.There is no restriction as to the number of prizes any boy gets.
III.No boy gets all the prizes?

Answer 1

Hint: The given question is related to permutations and combinations.Since each boy is different from other boys, we can say that the question is of permutation where it matters who gets the prize. So we will apply the formula for permutations, the formula for permutation is given by,
 $ {}^n{P_r} = \dfrac{{n!}}{{(n - r)!}} $
This is when the repetition is not allowed,
If the repetition is allowed, we can write the permutations as $ {n^r} $ .

Complete step-by-step answer:
Part I
The first part asks us to distribute the prizes to $ 6 $ boys, when no boy gets the prize twice,
This is the case of simple permutation so we can write this permutation as,
 $ {}^6{P_3} $ which can be solved as,
 $ {}^6{P_3} = \dfrac{{6!}}{{3!}} $
 $ {}^6{P_3} = \dfrac{{6 \times 5 \times 4 \times 3!}}{{3!}} $
 $ {}^6{P_3} = 6 \times 5 \times 4 $
Solving which we get,
 $ {}^6{P_3} = 120 $
Thus in the first part the prize can be distributed in $ 120 $ ways where no boy gets more than one prize.

Part II
The boys can have any number of awards, so each award can be distributed in $ 6 $ ways , since there are three awards we write,
The permutation will be given by,
 $ P = {n^r} $
 $ P = {6^3} $
 $ P = 216 $
Hence the award can be distributed in $ 216 $ ways.

Part III
The third part asks us to eliminate the cases where a person receives all the three awards; the rest of the cases will be the same as part $ 2 $ .
The cases where one person receives all the prizes will be $ 6 $ because there are $ 6 $ persons.
So the total cases will be
 $ 216 - 6 = 210 $
Hence the cases where no boy gets all the prizes will be $ 210 $ .

Note: The alternative way of doing the part second of the question would have been to just see that each award can be distributed to $ 6 $ persons. This process will be repeated $ 3 $ times, so the resultant number of permutations will be :
 $ P = 6 \times 6 \times 6 $
 $ P = 216 $
Which is the same as the figure we obtained in the question.