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Pardon my crudeness, as I'm not a mathematician. I have two unequal sets of items: $A = \{a,b,c,d,e,f\}$ and $B = \{1,2,3,4,5,6,7,8,9\}$. The members in each set can have multiple combinations. For example: $\{ab,abc,...\}$ and $\{1,12,123,...\}$. Order doesn't matter, so $\{ab,123\}$ is the same as $\{ba,312\}$.

To figure this out, I first simply added up the results from applying the binomial coefficient formula to each set and then multiplied the two sums:

${9}\choose{1}$ = $\frac{9!}{1!(9-1)!}$ + ${9}\choose{2}$ = $\frac{9!}{2!(9-2)!}$ + ... + ${9}\choose{9}$ = $\frac{9!}{9!(9-9)!}$ = 511

x

${6}\choose{1}$ = $\frac{6!}{1!(6-1)!}$ + ${6}\choose{2}$ = $\frac{6!}{2!(6-2)!}$ + ... + ${6}\choose{6}$ = $\frac{6!}{6!(6-6)!}$ = 63

The product of the two sums gave me 32,193 total combinations. Was my process correct? If not, what did I do wrong? If I am right, is there a more efficient formula?

dnaeye
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  • is ${, 1}$ an answer? (nothing from the letters, a member from the numbers)? If not you have the correct answer. It may be useful to recall that the sum $\sum_{k=0}^n {n \choose k} = 2^n$ – David Diaz Sep 28 '20 at 01:52
  • It seems that you are correct, if you don't allow the selection to be empty. – Lynnx Sep 28 '20 at 01:52
  • Thanks, Davids! I didn't know about that formula. I love learning new things about math! – dnaeye Sep 28 '20 at 02:55

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If you don't allow selecting nothing, then your answer is correct, but there is a quicker method. You have $2$ options on $a$, you can choose to select it or not to. And the same goes for others. Then you need subtract $1$ for the empty selection, giving:

Set $A$: $2^6-1=63$

Set $B$: $2^9-1=511$

Total: $63\cdot 511=32193$

Lynnx
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