Suppose six men and six women need to stand in a line at the bank. Assume that a

Question

Suppose six men and six women need to stand in a line at the bank. Assume that all the twelve are distinct human beings.

(a) In how many ways can it be done, if there are no restrictions ?

(b) In how many ways can it be done if the men and women must alternate ? (hint: stand the men first in their places).

(c) In how many ways can it be done if all the men must stand next to each other ? (hint: consider the group of men as one big blob first).

(d) In how many ways can it be done if at least one of the first two in the line must be a woman ?

(e) Suppose A and B are two specific people (out of the twelve). In how many ways can all the people stand in a line so that A and B are not next to each other ?

(f) Suppose A, B, C, and D are four specific people (out of the twelve). In how many ways can all the people stand in a line so that either A and B are next to each other or C and D are next to each other ?

Explanation / Answer

a)

If all are distinct and 12 are statding in a row the number of ways they can stand is= 12!= 479001600

b)

If they are standing alternative the number of ways will be= 6!*6!= 518400,

because we can arrange men who are 6 at alternate position by making position of women constant, so the number of ways will be 6!, now at each 6! times women starts shuffling so they will shuffle in 6! ways.

c)

Now if all men are standing together let's consider them as a whole single unit and then calculate, so the arrangements will be in 7 units(6 women and 1 unit of pack of men)and number of ways will be= 7!= 5040

d)

In that case the scenario will be in the first two:

So we will calculate for each case and sub them up.

1. the other 10 members can be shuffled in 10! ways= 3628800

2. same goes here 10! ways= 3628800

3. Same goes here 10! ways= 3628800

So 10886400 ways are possible.

e)

If A and B are not next to each other then men can shuffle (4!+4!+4!+4!) ways,

and women can shuffle completely, so that is 6!

So outcome will be= 4*4!*6!= 69120

f)

If A and B are together 11! and if C and D are together 11!

So 11!+11!= 79833600.

I hope this helps. Don't forget to give a thumbs up if you like this.

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