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everyone. I'm faced with a problem that I cannot solve without error.

There are $4$ blanks. Each one of those blanks has a possibility of different letters/numbers. Here's the full problem.

Blank one: $C,0,Q$

Blank two: $3,2$

Blank three: $S,9$

Blank four: $D,B,8$

Only one from each set can be selected. How can I get all of the combinations?

Example: $\{C,3,S,D\},\{0,3,S,D\} ...$

Anne Bauval
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Bowser5543
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    Suppose you have the first two blanks only. For blank 1 you can select from 3 different letters. For blank 2 you can select from 2 different numbers. In total you have $3\cdot 2 =6$ possible combinations. It is comprehensible ? – callculus42 Aug 24 '15 at 06:12
  • @calculus That makes sense. Would that mean there would be 36 combinations, then? I'm assuming not.. It would appear that the amount of possibilities would be much higher.. But if that isn't the case, how can I successfully calculate all of the possibilities without repeats? – Bowser5543 Aug 24 '15 at 06:16
  • You are right. If you cannot believe it, you can write them down. And there are no repeats, because in each blank you can put only one letter/number. – callculus42 Aug 24 '15 at 06:20
  • @calculus Alright. If 36 total combinations are possible, it's not too hard to write all of them. In reference to repeats, I'm trying to write this on paper without accidently putting the same one twice. (I have pretty low memory.) I'll start now. Stopping at 36. – Bowser5543 Aug 24 '15 at 06:22
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    If you write them on paper systematically, then it should not happen,that you repeat a combination. But nice that you wrote them down. It is one important way to get a better understanding of combinatoric. – callculus42 Aug 24 '15 at 06:29
  • @calculus Thanks. I'll do my best and report the results. – Bowser5543 Aug 24 '15 at 06:31
  • Let us continue this discussion in chat. – Bowser5543 Aug 24 '15 at 06:34

2 Answers2

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The set of all possible outcomes $$= \{(C,3,S,D),(C,3,S,B),(C,3,S,8),(C,3,9,D),(C,3,9,B),(C,3,9,8),(C,2,S,D),(C,2,S,B),(C,2,S,8),(C,2,9,D),(C,2,9,B),(C,2,9,8),(o,3,S,D),(o,3,S,B),(o,3,S,8),(o,3,9,D),(o,3,9,B),(o,3,9,8),(o,2,S,D),(o,2,S,B),(o,2,S,8),(o,2,9,D),(o,2,9,B),(o,2,9,8),(Q,3,S,D),(Q,3,S,B),(Q,3,S,8),(Q,3,9,D),(Q,3,9,B),(Q,3,9,8),(Q,2,S,D),(Q,2,S,B),(Q,2,S,8),(Q,2,9,D),(Q,2,9,B),(Q,2,9,8)\}.$$

James Pak
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Well, actually this is a very simple introductory problem to Combinations,

You have $3$ possible combinations for the $1$st blank so you get $\binom31$ = $3$
Also for the $2$nd blank we have $2$ options so you get $\binom21$ = $2$
Similarly, for the $3$rd blank you have $2$ options so you get $\binom21$ = $2$
And for the last blank you have $3$ options so you get $\binom31$ = 3

Now multiply all the combinations you get, $2\times2\times3\times3$ = $2^2\times3^2$ = $36$ which is the answer you get by listing without listing and wasting time.

Prakhar Nagpal
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