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Question: The number of ways in which first, second and third prizes can be given to \(5\) competitors is \...

The number of ways in which first, second and third prizes can be given to 55 competitors is
A)10A)10
B)60B)60
C)15C)15
D)125D)125

Explanation

Solution

We make use permutation and combination methods to approach the given questions to find the number of ways since the number of permutations of r-objects can be found from among n-things is npr{}^n{p_r}where p refers to the permutation. Also, for combination, we have r-things, and among n-things are ncr{}^n{c_r}.
In this problem, they asked us to find possible ways for the first, second and third prizes to be given to 55 competitors. Hence, we use the permutation method.
Formula used:
Permutation formula is npr=n!(n−r)!{}^n{p_r} = \dfrac{{n!}}{{(n - r)!}}, where n is the total objects count and r is the possible ways.

Complete step-by-step solution:
Given that we have, the number of ways in which first, second and third prizes. Thus, we have three possible outcomes. The total objects are 55 competitors.
Hence in the five competitors, we need to distribute the price to three members in the order of any first, second, and third prices.
Hence the number of possible ways is npr=n!(n−r)!⇒5p3=5!(5−3)!{}^n{p_r} = \dfrac{{n!}}{{(n - r)!}} \Rightarrow {}^5{p_3} = \dfrac{{5!}}{{(5 - 3)!}} where the total objects are five and the possible outcomes are three.
Further solving we get 5p3=5!(5−3)!⇒5!2!=5.4.3.2.12.1⇒5×4×3{}^5{p_3} = \dfrac{{5!}}{{(5 - 3)!}} \Rightarrow \dfrac{{5!}}{{2!}} = \dfrac{{5.4.3.2.1}}{{2.1}} \Rightarrow 5 \times 4 \times 3 (where factorial represented as n!=n(n−1)(n−2)....2.1n! = n(n - 1)(n - 2)....2.1)
Hence, we get 5p3=5×4×3⇒60{}^5{p_3} = 5 \times 4 \times 3 \Rightarrow 60 (with the help of multiplication operation)
Therefore, the option B)60B)60 is correct.

Note: We are also able to solve this problem by using the permutation and combination method.
Since there are a total of five prizes and the first prize can be distributed in the 55 ways.
Since one person got the first prize so he will not be repeated for the second and third prize, thus the second prize can be distributed in 5−1=45 - 1 = 4 ways.
Similarly, for the third prize (first and second prize people will not be repeated) we have 5−1−1=35 - 1 - 1 = 3 ways.
Hence in total, we have 5.4.3⇒605.4.3 \Rightarrow 60 and hence the result.