3
$\begingroup$

Please help me in understanding the difference between this two question:

How many ways can $4$ men and $4$ women stand in a line if the women and men alternate ?

How many ways can $4$ men and $4$ women stand in a line if no two women are together ?

We employ a bit of different strategies for solving them and the answer is also different,but I am not understanding why is this differences.Kindly explain.


I am in bit of a doubt in another similar problem:

Find the number of ways of arranging $21$ white balls and $3$ black balls in a row so that no two black balls are together.

I think the answer should be $21! \times ^{24}P_3$ but this is incorrect.Could somebody explain?

$\endgroup$

1 Answer 1

7
$\begingroup$

In the first point, the problems are different because the second problem would allow e.g. WMW*MM*WMW while the first would not. (The first is essentially saying "no two women are together and no two men are together", while the second only says "no two women are together".)

In the second point, I think you're making two mistakes:

  • By multiplying by $21!$, and by using $P$ instead of $C$, you're assuming you can tell the difference beween the white balls, and can tell the difference between the black balls. (i.e. B1 B2 W1 W2... is different from B2 B1 W1 W2...) Usually when dealing with balls, they're indistinguishable.

  • The answer $^{24}C_3$ (i.e. choosing any 3 balls out of the 24 to be black) would be incorrect because it doesn't exclude answers such as BWBBW... where two black balls appear next to each other.

$\endgroup$
4
  • $\begingroup$ Thanks I did understood your first point but I don't understand your second point. $\endgroup$
    – Quixotic
    Dec 21, 2010 at 12:00
  • $\begingroup$ @Tretwick I've expanded the second point a bit. Is that any clearer? $\endgroup$
    – Rawling
    Dec 21, 2010 at 12:08
  • $\begingroup$ Thanks,now I understand where I was wrong,I realized that the answer should be $^{22}C_3$.A possible explanation is: If we arrange the 21 black balls first then in order to have a white ball between two black balls we have 22 cases. $\endgroup$
    – Quixotic
    Dec 21, 2010 at 14:18
  • $\begingroup$ That took me a minute to figure out, but yes, I think that's right. 21 white balls -> 22 slots between white balls -> choose three of these to put a black ball in. $\endgroup$
    – Rawling
    Dec 21, 2010 at 14:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .