Question

Number of ways in which 3 persons throw a normal die to have a total score of 11 is?

Answer 1

Answer:

27 WAYS

Step-by-step explanation:

Let us consider 3 persons are A, B and C.

Now, the possible values that A can get is {1,2,3,4,5,6} and is same with B and C.

We have to get the total of 11. Check, in how many possible ways we get total 11 by using {1,2,3,4,5,6} values.

The possible sets are:

{6,1,4}, {6,2,3}, {5,1,5}, {5,2,4}, {5,3,3}, {4,3,4}

Now, each possible set we can get in different ways.

{6,1,4} we can arrange this set in 3! i.e, 6 ways.

[{6,1,4}, {6,4,1}, {4,6,1}, {4,1,6}, {1,4,6}, {1,6,4}].

In the same lines we can arrange each possible set in 3! ways if there are no repeats in the set.

If there are any repeats in set, consider {5,3,3} we can arrange this possible set by (3!/2!) i.e, 3 ways.

[{5,3,3}, {3,5,3}, {3,3,5}]

Now, the total possible sets to get sum = 11 are,

3! + 3! + (3!/2!) + 3! + (3!/2!) + (3!/2!) = 27 ways.