Question

In how many ways can 4 prizes each having 1st,2nd and 3rd positions be given to three boys, if each boy is eligible to receive one prize for each event?
A). $^{12}{P_3}$
B). ${6^4}$
C). ${4^3}$
D). ${4^6}$

Answer 1

In the given question, we have been given that there are some given number of items. We have to make some given number of choices for the given items. To do that, we use the formula of factorial for distributing some $n$ things among $m$ people with replacement. Then we take the number of cases and raise it to the power of the factorial calculated.

Formula used:
We have to calculate the value of the possibilities using the formula of factorial,
$n! = n \times \left( {n - 1} \right) \times \left( {n - 2} \right) \times ... \times 2 \times 1$

Complete step by step solution:
We have to give the prizes to $3$ boys.
For each prize, there are three positions, hence, the number of ways of arranging $3$ things at $3$ positions is:
$3! = 3 \times 2 \times 1 = 6$
Now, there are $4$ prizes, thus the total number of ways is:
${6^4} = 1296$
Hence, the correct option is B.

Note: In the given question, we had to calculate the number of ways of giving an entity to some given number of other entities, with replacement. We did that by using the formula of factorial and then raising the latter entity to the power of factorial. We just need to remember all the formulae and know their basic meaning about when one is used and when the other is used.