Question

The number of ways in which we can get a score of 11 by throwing three dice is
(a) 18
(b) 27
(c) 45
(d) 56

Answer 1

In order to solve this question, we will find different cases for 2 dice for different values of 3 die. Also, we have to remember that the maximum possible value of each die is 6 and the minimum possible value of each die is 1. By using this, we can find the answer to this question.

Complete step-by-step answer:
In this question, we have been asked to find the number of ways in which we can get a score of 11 by throwing 3 dice. For that, we can write the condition as, ${{x}_{1}}+{{x}_{2}}+{{x}_{3}}=11$ where $1\le {{x}_{1}},{{x}_{2}},{{x}_{3}}\le 6$

Now, we will consider different values of ${{x}_{1}}$ and for that we will find the values of ${{x}_{2}}$ and ${{x}_{3}}$ and then we will be able to find the number of ways of getting a score of 11. Let us consider, we get ${{x}_{1}}=1$. So, the possible values for ${{x}_{2}}+{{x}_{3}}=10$ will be (4, 6), (5, 5) and (6, 4) which gives us 3 ways.

Now, for ${{x}_{1}}=2$, we will get the values of ${{x}_{2}}+{{x}_{3}}=9$ by (3, 6), (4, 5), (5, 4), (6, 3) as 4 combinations only.

Now, for ${{x}_{1}}=3$, we will get the values of ${{x}_{2}}+{{x}_{3}}=8$ by (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) as 5 possible combinations only.

Now, for ${{x}_{1}}=4$, we will get the values of ${{x}_{2}}+{{x}_{3}}=7$ by (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1) as 6 possible combinations.

Now, for ${{x}_{1}}=5$, we will get the values of ${{x}_{2}}+{{x}_{3}}=6$ by (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) as 5 possible combinations.

Now, for ${{x}_{1}}=6$, we will get the values of ${{x}_{2}}+{{x}_{3}}=5$ by (1, 4), (2, 3), (3, 2) and (4, 1) as 4 possible combinations.

So, for the total number of cases, we will add up all these values, therefore we get,
3 + 4 + 5 + 6 + 5 + 4 = 27 ways
Therefore, there are 27 ways for getting a score of 11 when 3 dice will be thrown.

Hence, option (b) is the right answer.

Note: We can also solve this question by finding the power of ${{x}^{11}}$ in ${{\left( x+{{x}^{2}}+{{x}^{3}}+{{x}^{4}}+{{x}^{5}}+{{x}^{6}} \right)}^{8}}$. We can see that it will give us the most appropriate answer. Also, we have to be very careful while solving this question because of lots of consideration. We must not miss out any of the possible combinations.