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I am creating a custom product that customers can select parts and build their own end product. To be specific, it is a costume sword made up of the following parts: blade, guard, grip, and pommel. I have varying options for each piece, such as follows:

5 pommels

10 grips

7 guards

4 blades

Obviously, only $1$ of each part can be selected (you can't build a sword with $2$ blades), and one option MUST be selected from each part (You can't build a sword WITHOUT a blade).

I would like to advertise how many possible combinations there are, but I am a creative person, not a mathematical one, and trying to ensure I am accurate is a tall order... if I am correct, the formula would be $$\binom{5}{1}\binom{10}{1}\binom{7}{1}\binom{4}{1} = 1400$$

Is this correct?

Thank you in advance!

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  • $\begingroup$ Here is a tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig May 24 '17 at 8:56
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Your answer is correct.

The Multiplication Principle states that if one task can be performed in $m$ ways and a second task can be performed independently of the first in $n$ ways, then both tasks can be performed in $mn$ ways. Hence, the number of ways of selecting a pommel, a grip, a guard, and a blade from $5$ pommels, $10$ grips, $7$ guards, and $4$ blades is $5 \cdot 10 \cdot 7 \cdot 4 = 1400$.

Note that $$\binom{n}{k} = \frac{n!}{k!(n - k)!}$$ and that $$\binom{n}{1} = \frac{n!}{1!(n - 1)!} = \frac{n(n - 1)!}{1!(n - 1)!} = \frac{n}{1} = n$$ so $$\binom{5}{1}\binom{10}{1}\binom{7}{1}\binom{4}{1} = 5 \cdot 10 \cdot 7 \cdot 4 = 1400$$

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