In how many ways can a score of $15$ or $16$ be obtained if a player throws $6$ darts at a board with possible scores of $0$, $1$, $2$ or $3$?

A dart board is constructed. The scores you can get are $0$, $1$, $2$ or $3$. Bob and Lily have a game where each player throws $6$ darts at the board. The scores gained from the $6$ throws are added. In how many ways can a total score of 15 or 16 be gained?

Note The total score gained from $2+0+0+3+3+2$ is different from the total score gained from $0+0+2+3+3+2$

From my calculations, there should be $77$ ways. Is this correct?

• Yes, your answer seems absolutely correct! Jul 3 '17 at 6:41

Your answer is correct. Since you have not explained how you arrived at your answer, here is one method. Jonathan Davidson has provided another using a recursion.

In order to obtain a score of $16$, either five 3's and one 1 are obtained or four 3's and two 2's are obtained.

1. five 3's and one 1: There are $\binom{6}{5}$ ways for five of the six throws to be 3's.
2. four 3's and two 2's: There are $\binom{6}{4}$ ways for four of the six throws to be 3's.

Total: Since the above cases are disjoint, the number of ways of obtaining a score of 16 is $$\binom{6}{5} + \binom{6}{4} = 6 + 15 = 21$$

In order to obtain a score of 15, either five 3's and one 0 are obtained; four 3's, one 2, and one 1 are obtained; or three 3's and three 2's are obtained.

1. five 3's and one 0: There are $\binom{6}{5}$ ways for five of the six throws to be 3's.
2. four 3's, one 2, and one 1: There are $\binom{6}{4}$ ways for four of the six throws to be 3's. There are $\binom{2}{1}$ ways to fill the leftmost open position. Hence, there are $\binom{6}{4}\binom{2}{1}$ possible sequences in this case.
3. three 3's, three 2's: There are $\binom{6}{3}$ ways for three of the six throws to be 3's.

Total: Since the above cases are disjoint, the number of ways of obtaining a score of $15$ is $$\binom{6}{5} + \binom{6}{4}\binom{2}{1} + \binom{6}{3} = 6 + 15 \cdot 2 + 20 = 6 + 30 + 20 = 56$$

Hence, the number of ways of obtaining a score of 15 or 16 is $56 + 21 = 77$, as you found.

Let $N_k(s)$ denote the number of ways $\sum_{j=1}^k x_j = s$ where $x_j$ is the score of the $j$th dart. Then \begin{align} N_6(16) &= N_5(13)+N_5(14)+N_5(15) \\ &= N_4(10)+2N_4(11)+3N_4(12) \\ &= N_3(7)+3N_3(8)+6N_3(9) \\ &= N_2(4)+4N_2(5)+10N_2(6) \\ &= N_1(1)+5N_1(2)+15N_1(3) \\ &= 21 \end{align} A similar calculation shows $N_6(15) = 56$. Therefore, there are 77 ways in which a score of 15 or 16 can be achieved.

• what would the calculation for N6(15) be? Jul 3 '17 at 7:27
• It would be 56. I do not want to type up the calculation. The Idea behind my process is the recurrence $$N_k(s) = N_{k-1}(s-3)+N_{k-1}(s-2)+N_{k-1}(s-1)+N_{k-1}(s)$$ Jul 3 '17 at 7:30
• A suggestion: Place the recursion you are using in your answer rather than the comments. Jul 3 '17 at 9:09

You are allowed to loose $2$ or $3$ points. The partitions of $2$ are $(2)$ and $(1,1)$, which can be realized in $6+{6\choose2}=21$ ways. The partitions of $3$ are $(3)$, $(2,1)$, and $(1,1,1)$, which can be realized in $6+6\cdot5+{6\choose3}=56$ ways. Makes $77$ in total.