# Permutations and combinations two part question

This is for some personal combinations / permutations study, I was wondering about a certain type of question that I shall phrase thus:

Given 17 of object A and 13 of object B, how many ways may four A objects and three B objects be ordered in a line?

I'm uncertain as to how this question may be answered. Obviously 4C17 is the number of ways four object A's may be ordered, and 3C13 the same for three object B's, yet how may one combine the two?

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It depends on whether the objects are distinguishable. I’ll assume for now that they are. First choose which $4$ of the $7$ places in the line are to be occupied by $A$ objects; this can be done in $\binom74$ ways. (By the way, if you use the $C$ notation, that’s $^7C_4$, not $^4C_7$.) Now you have $17$ ways to pick the first $A$ object, $16$ ways to pick the next $A$ object, $15$ ways to choose the third, and $14$ ways to choose the last $A$ object. After that you can choose the first $B$ object in $13$ ways, the second in $12$ ways, and the last in $11$ ways. The total number of lineups is therefore

$$\binom74\cdot17\cdot16\cdot15\cdot14\cdot13\cdot12\cdot11=3,430,627,200\;.$$

Added: As Henry points out in the comments, there is another straightforward way to analyze this version of the problem. There are $\binom{17}4$ ways to choose $4$ $A$ objects, $\binom{13}3$ ways to choose $3$ $B$ objects, and then $7!$ possible orders for the $7$ objects chosen, for a total of $$\binom{17}4\binom{13}37!$$ lineups.

If the various $A$ objects are indistinguishable from one another, as are the $B$ objects, then the only thing that distinguishes one lineup from another is the placement of the $A$ objects in the line, so there are only $\binom74=35$ lineups.

If the $A$ objects and the $B$ objects are distinguishable, so that you can tell which $4$ of the one you have and which $3$ of the other, but you don’t care about the order in which the $A$ objects occupy the $4$ $A$ slots or the order in which the $B$ objects occupy the $3$ $B$ slots, but you do care which slots in the lineup are filled with $A$ objects and which with $B$ objects, then you $\binom74$ ways to choose which $4$ slots get $A$ objects, $\binom{17}4$ ways to choose which $4$ $A$ objects to use, and $\binom{13}3$ ways to choose the $3$ $B$ objects, for a total of

$$\binom74\binom{17}4\binom{13}3=23,823,800$$

lineups.

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Your first expression is also $7! {17 \choose 4}{13 \choose 3}$ –  Henry Sep 18 '12 at 8:58
@Henry: True; I didn’t even notice that route, but it’s well worth adding. –  Brian M. Scott Sep 18 '12 at 9:03