Question

The number of ways in which 13 gold coins can be distributed among three persons such that each one gets at least two gold coins is
A. 36
B. 24
C. 12
D. 6

Answer 1

Hint: We must distribute 13 coins such that each one gets at least two. So, we can give the 3 persons 2 coins each and then distribute the rest of the coins to the 3 persons. So, the required number of ways is the number of ways of distributing the rest of the coins among 3 persons. The number of ways of distributing n objects among r objects is given by, ${}^{n + r - 1}{C_{r - 1}}$. We can simplify this using the equation ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$

Complete step by step Answer:

We have 13 gold coins and 3 persons. Each one must get at least 2 coins. So, we can give each person 2 coins each.
Then the number of coins that are left is $13 - 2 \times 3$
$ = 13 - 6 = 7$
Therefore, there are 7 coins left. These can be distributed among the 3 persons without any
conditions.
Thus, the number of ways of distributing 13 coins among 3 persons such that each one gets at least two coins is equal to the number of ways of distributing 7 coins among 3 persons.
The number of ways of distributing n objects among r objects is given by, ${}^{n + r - 1}{C_{r - 1}}$.
We can substitute the value of n and r to the equation.
$ \Rightarrow {}^{7 + 3 - 1}{C_{3 - 1}} = {}^9{C_2}$
We know that ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$.
$ \Rightarrow {}^9{C_2} = \dfrac{{9!}}{{2!\left( {9 - 2} \right)!}}$
On simplification, we get,
$ \Rightarrow {}^9{C_2} = \dfrac{{9!}}{{2!\left( 7 \right)!}}$
We know that, $n! = n \times \left( {n - 1} \right)!$. Applying this condition on the numerator, we get,
$ \Rightarrow {}^9{C_2} = \dfrac{{9 \times 8 \times 7!}}{{2!\left( 7 \right)!}}$
On cancelling common terms we get,
$ \Rightarrow {}^9{C_2} = \dfrac{{9 \times 8}}{2}$
$ \Rightarrow {}^9{C_2} = 36$
Thus, the number of ways of distributing 7 coins among 3 persons is 36.
Therefore, the number of ways of distributing 13 coins among 3 persons such that each one gets at least two coins is also 36.
Therefore, the correct answer is option A.

Note: We used the concept of combination to find a number of ways. Here, all the coins are identical and the order is not important. As each person has to get at least 2 coins, we only need to calculate the combination of dividing the rest of the coins with the 3 persons. We must take care while simplifying the factorials and other calculations. The factorial of a positive integer is defined as the product of all the positive integers less than or equal to that integer. It is represented as $n!$