The number of ways in which a score of $11$ can be made from a throw by three persons,each throwing a single die once is

Question

What is the number of ways in which a score of $11$ can be made from a throw by three persons, each throwing a single die once?

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My Attempt:

I tried to answer this question by counting the possible number of occurences of dices such that their sum is $11$ without repetition of same occurrences.

$6$ $4$ $1$
$6$ $3$ $2$
$5$ $5$ $1$
$5$ $4$ $2$
$6$ $3$ $3$
$4$ $3$ $4$

So the total number of such occurrences is $6$ and these occurrences can be permuted in $3!$ ways each. So the final answer is $6 \cdot 3!$. But the correct answer is supposed to be 27, so where am I wrong?

Note: I want to solve this question by counting method only.

Answer 1

You are counting the permutations of $434$, $551$ and $633$ as six possibilities each, rather than the correct three. Thus, you must subtract nine from your total of $36$ to get the correct result of $27$.

Answer 2

You can also write every possible ways in which 11 comes keeping in mind that not repeating the digits. For example if i have counted (1,4,6) i must not count (6,4,1) and other nos. That can be formed using (1,4,6) while writing down the ways in which no. Adds up to 11. ( We will consider only when counting the total no. of possibilities.) (1,4,6)→6 possibilities (1,5,5)→3 possibilities (2,3,6)→6 possibilities (2,4,5)→6 possibilities (3,3,5)→3 possibilities (3,4,4)→3 possibilities In total 27 possibilities.

Answer 3

By using generating functions, we are interested in the coefficient of $t^{11}$ in the expression

$$(t+t^2+t^3 + t^4 + t^5 + t^6)^3$$

which is $27$.