arrangement of a group of people.

Question

The no. of ways in which 4 particular persons A,B,C,D and 6 other persons can stand in a queue so that A always stands before B, B before C and C before D is ?

My try: since A B C D always have to stand together taking them as one, we get, 7! Ans

Is this correct?. Help needed.

score 2 · Answer 1 · answered Nov 05 '13 at 13:40

The question doesn't state that A, B, C, and D stand together, only that A is before B (there may be people in between), and B before C, and C before D.

Hint: Consider all $10!$ ways of lining up 10 people. What fraction of them have A before B, B before C, and C before D?

Librecoin · Answer 2 · 2013-11-05T17:37:48.627

2

First, choose $4$ out of $10$ places for the persons $A$, $B$, $C$, $D$. This can be done in $n_1={10\choose4}=\frac{10!}{4!(10-4)!}$ ways.

Then, distribute the other persons into the remaining $6$ places. This can be done in $n_2=6!$ ways.

Finally, by the principle of multiplication, the answer is given by $n_1n_2=\frac{10!}{4!}$.

edited Nov 05 '13 at 17:37

answered Nov 05 '13 at 13:53

Librecoin

2,740

1

Also, the answer has to be $\dfrac{10!}{4!}$ because, if the $10!$ arrangements are equally likely, the probability of $A,B,C,D$ being in that particular order is $\dfrac1{4!}$. – bof Nov 05 '13 at 14:18

score 1 · Answer 3 · answered Nov 05 '13 at 13:41

1

It depends on what you mean by "stand before." If you take it to mean immediately before, then your $7!$ is correct. But if it just means that $A$ is somewhere ahead of $B$, etc., then you get a different answer. Can you take it from here?

answered Nov 05 '13 at 13:41

Barry Cipra

79,832

Got it... thanks – john Nov 05 '13 at 13:47

arrangement of a group of people.

3 Answers3