# In a shooting a man scored 5,4,3,2,0 points for each slot ,then in how many ways can the person score 30 in seven shots?

In a shooting a man scored 5,4,3,2,0 points for each slot ,then in how many ways can the person score 30 in seven shots ?

Firstly ,can someone please explain why this is a combinations question and not a permutations question .

• As a way to get started, first argue that you need at least two $5's$. Place those, now you are down to five shots that sum to $20$. – lulu Apr 20 at 12:19
• The problem is not well-defined. If a shooter shoots 5,5,5,5,5,5,0, is that different from 5,5,5,5,5,0,5? The number of ways the shooter can shoot 6x 5-pt shots and 1x 0-pt shots is $$\dbinom{7}{1}(1)^6(1)^1$$ which uses binomial coefficients as the problem desires (it is the binomial formula, in fact). But it is a calculation of permutations. If it is the same, then you want the number of solutions to the system of Diophantine equations: $$\begin{matrix}5x_1 & + & 4x_2 & + & 3x_3 & + & 2x_4 & +& 0x_5 & = & 30 \\ x_1 & + & x_2 & + & x_3 & + & x_4 & + & x_5 & = & 7\end{matrix}$$ – InterstellarProbe Apr 20 at 12:22
• ... in nonnegative integers. – InterstellarProbe Apr 20 at 12:24
• Sorry i don't know what diophantine equations are ..is there another way to solve the system of equations ? – Tanya Apr 20 at 12:27
• Yes, many ways. Are you saying the problem considers the man shooting 6x 5-pt shots and 1x 0-pt shot as a single solution rather than seven solutions? – InterstellarProbe Apr 20 at 12:30

Suppose you have 2x 0-pt shots. That leaves 5 shots to make 30 point. The maximum score in 5 shots is 25, so that is not possible. You can have a maximum of 1x 0-pt shot. If you do, you have to have all the rest as 5-pt shots.

So, there is 1 solution where you have a 0-pt shot. The rest are solutions with no 0-pt shots. So, we consider the problem again with the only possible shots as 5, 4, 3, or 2 pt shots.

Suppose we have 2x 2-pt shots. That leaves 5 shots to make 26 points, which is not possible. So, you have either 0 or 1 2-pt shot.

Consider the case with 1 2-pt shot: That leaves 28 points for 6 shots with possible point values of 5,4,3. If you have 1x 3, the rest must be 5's. Otherwise, you are looking to make 28 points with only 4's and 5's. There is only one way to do that. So, this case has 2 solutions (2,3,5,5,5,5,5 and 2,4,4,5,5,5,5)

With 0 2-pt shots: That leaves 30 points for 7 shots with possible point values of 5,4,3. You can have 0, 1, or 2 3-pt shots. Any more than 2 and you will be unable to make 30 points in the remaining shots. If you have 2x 3-pt shots, you have 5 shots remaining with only 4-pt and 5-pt shots, and you must make 24 points. That can only be done with 3,3,4,5,5,5,5

If you have 1x 3-pt shot, that leaves 27 points in six shots with only 4-pt and 5-pt shots. 4x+5y=27 and x+y=6 is a simple system of equations to solve. It has only one solution.

Finally, if you have 0x 3-pt shots, that leaves 30 points in seven shots with only 4-pt and 5-pt shots. 4x+5y=30 and x+y=7 is a simple system of equations to solve that has only one solution.

So, here is a complete list of all possible ways 30 points can be achieved: (0,5,5,5,5,5,5) (2,3,5,5,5,5,5) (2,4,4,5,5,5,5) (3,3,4,5,5,5,5) (3,4,4,4,5,5,5) (4,4,4,4,4,5,5)

Because we exhausted every possible case, there cannot be any other solutions.

Now, if you want the number of ways the man can achieve these shot distributions (finding the number of orders), we find the number of orders of each case. This is multiset permutations.

(0,5,5,5,5,5,5): $$\dfrac{7!}{1!6!} = 7$$

(2,3,5,5,5,5,5): $$\dfrac{7!}{1!1!5!} = 42$$

(2,4,4,5,5,5,5): $$\dfrac{7!}{1!2!4!} = 105$$

(3,3,4,5,5,5,5): $$\dfrac{7!}{2!1!4!} = 105$$

(3,4,4,4,5,5,5): $$\dfrac{7!}{1!3!3!} = 140$$

(4,4,4,4,4,5,5): $$\dfrac{7!}{5!2!} = 21$$

So, the total number of ways he can make those shots (if order of his shots matters) would be 420. If order of his shots does not matter, the answer is 6.

The result of 420 can also be achieved by finding the coefficient of $$x^{30}$$ in the expansion of $$(1+x^2+x^3+x^4+x^5)^7$$: Wolframalpha

In other words, if order of the shots matters, the problem is significantly easier.

• Thankyou so much .But why will the result of 420 be achieved by finding the coefficient of x30 in the expansion ? – Tanya Apr 20 at 13:38
• As an example of how this works, one term that will get added together to become the coefficient of $x^{30}$ would be $x^0$ (from the first term) and $x^5$ from every other term. So, the coefficient of $x^{30}$ is calculated like this: $$(x^0)(x^5)(x^5)(x^5)(x^5)(x^5)(x^5) + (x^2)(x^3)(x^5)(x^5)(x^5)(x^5)(x^5)+\cdots+(x^5)(x^5)(x^5)(x^5)(x^5)(x^5)(x^0)$$ Do the exponents look similar to the calculations above? – InterstellarProbe Apr 20 at 14:00
• Is there a name to this concept ?Or a general rule ? – Tanya Apr 20 at 14:32
• Look up the Multinomial Theorem, specifically the different Interpretations. This is a method of counting the number of ways to put distinct objects into distinct bins. – InterstellarProbe Apr 20 at 16:14
• @Tanya The expression $(1+x^2+x^3+x^4+x^5)^7$ is a generating function for the sequence in which $s_n$ is equal to the number of ways to score $n$ in 7 shots with the given shot values. – amd Apr 20 at 19:12