Question

If 4 dies is rolled, then the number of ways of getting the sum is equal to 10 is

& \text{(A) 56} \\\ & \text{(B) 64} \\\ & \text{(C) 72} \\\ & \text{(D) 80} \\\ \end{aligned}$$

Answer 1

We know that the number of positive integral solutions of ${{\text{x}}_{1}}+{{x}_{2}}+......+{{x}_{r}}=n$ is equal to $^{n-1}{{C}_{r-1}}$. Now by using this concept, we should find the number of ways to have the total sum is equal to 10 if 4 dies are rolled. Now according to the problem, we should mention the type of outcomes.

Complete step-by-step answer:
In a die, we can have 6 outcomes. We can get 1, 2, 3, 4, 5 and 6 as outcomes in a die. In the question, it was given that 4 dies are rolled. Let us assume the outcomes on the 4 dies are ${{\text{x}}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}}$ respectively. We were given that the sum of all the 4 outcomes on the four dies is equal to 10.
So, we can write that
${{\text{x}}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}=10....(1)$
In equation (1), we can take the values of ${{\text{x}}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}}$ such that the repetition is allowed. The repetition is allowed because the value of die obtained can be repeated.
We know that the number of positive integral solutions of equation ${{x}_{1}}+{{x}_{2}}+{{x}_{3}}+......+{{x}_{r}}=n$ is equal to $^{n-1}{{C}_{r-1}}$.
From equation (1), we have to find the number of solutions of ${{\text{x}}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}=10$.
Now we have to compare ${{\text{x}}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}=10$ with ${{x}_{1}}+{{x}_{2}}+{{x}_{3}}+......+{{x}_{r}}=n$, we get the value of r is equal to 4 and the value of n is equal to 10.
Let us assume the number of solutions of ${{\text{x}}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}=10$ is equal to S.

& \Rightarrow S{{=}^{10-1}}{{C}_{4-1}} \\\ & \Rightarrow S{{=}^{9}}{{C}_{3}} \\\ & \Rightarrow S=\dfrac{9!}{6!3!} \\\ & \Rightarrow S=84 \\\ \end{aligned}$$ So, the number of solutions of $${{\text{x}}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}=10$$ is equal to 84. We know that the values of $${{\text{x}}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}}$$ must be greater than or equal to 1 and less than or less than 6. So, we have to exclude the number of ways formed by the numbers 7, 1, 1, 1 to make the sum. The number of ways to place 7, 1, 1, 1 in 4 places is equal to 4. So, the total number of ways $$=84-4=80$$. Hence if 4 dies is rolled, then the number of ways of getting the sum is equal to 10 is 80. So, option D is correct. **So, the correct answer is “Option D”.** **Note:** Students may forget to exclude the number of ways to place the values of $${{\text{x}}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}}$$ as 7,1,1,1 to have the condition satisfied $${{\text{x}}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}=10$$. If this point is forgotten by the student, then he/ she will if 4 dies is rolled, then the number of ways of getting the sum is equal to 10 is 84. But there is no option to satisfy. So, students should be careful in these cases.