How many pairs of dance partners can be selected from a group of $12$ women and $20$ men ?

Ans given : $P(20, 12)$

Shouldn't the answer be $20 × 12$ as the pair can be selected from any of the $12$ women and for each women there are $20$ men to choose for.


Shouldn't the answer be 20 x 12 as the pair can be selected from any of the 12 women and for each women there are 20 men to choose for.

$20{\times}12$ counts the ways to select just $1$ m:f-pair from the group of $20$ males and $12$ females.

${}^{20}\mathrm P_{12}$ counts the ways to arrange the group into $12$ m:f-pairs (and $8$ male wallflowers).   All at once.

So the answer depends on how you interpreted the question.


You first select 12 men from possible 20, that can be done in $\binom{20}{12}$ ways. Now these 12 men have to be paired with the 12 women. Each pairing is simply a bijective function from the set of 12 men to the set of 12 women. Number of such bijective mappings is $12!$. So in all $$\binom{20}{12} \cdot 12!=P(20,12) \quad \text{ways}.$$

  • $\begingroup$ I still don't get the idea. All I could see are these pairs ,<W1,M1>,<W1,M2>,<W1,M3>,....<W1,My>,<W2,M1>,<W2,M2> .... <Wx,My> where x=12;y=20 Can you please tell why my answer is wrong? $\endgroup$ – Teja713 Mar 11 '17 at 22:30
  • $\begingroup$ @Teja714 $12\cdot 20$ counts the ways to select 1 pair from the group. ${}^{20}\mathrm P_{12}$ counts the ways to form 12 pairs at once. (Note: this is an old problem that assumes you can only form male-female pairs.) $\endgroup$ – Graham Kemp Mar 12 '17 at 0:56

There are n*(n-1)*(n-2)....(n-r+1) possibilities

(ie) 20 ways to select first pair

19 ways to select second pair

18 ways to select third pair


9 ways to select last pair ie (20-12+1)

Thus 20*19*18....*9 ways are there

which is nothing but P(20,12)


Let $W$ denote the set of women and $M$ denote the set of men. An arrangement in which each woman has exactly one male dance partner is the same thing as an injection $W \rightarrow M$. So compute:

$$|M^{\downarrow W}| = |M|^{\downarrow |W|} = \prod_{w = 0}^{|W| - 1}(|M|-w).$$

Notation. whenever $A$ and $B$ are sets, the notation $A^{\downarrow B}$ means the set of all injections $B \rightarrow A,$ and whenever $a$ and $b$ are natural numbers, $a^{\downarrow b}$ denotes the falling factorial. Now just plug your numbers in.

  • $\begingroup$ Whoever downvoted my perfecty correct, and, in my opinion, fairly clear answer should be ashamed of themselves. $\endgroup$ – goblin Mar 12 '17 at 11:10

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