Please help me with the following questions:

A committee of 8 people consisting of 3 men and 5 women are lining up next to each other for a photograph.

i) In how many different ways can they be arranged for the photo?

ii) How many arrangements are possible if a male has to stand at both ends
of the line.

iii) In how many ways can a subcommittee of four people be chosen if it has to contain at least one male.

My attempt:

i) 8!/(3!5!) = 56 ways

ii) 6!/5! = 6 ways

iii) 8!/(5!3!) - 5!/(2!4!) = 54 ways

But I am not confident with my answers at all.

  • $\begingroup$ II should be less than i! $\endgroup$ – herb steinberg Jun 20 '19 at 17:20
  • 1
    $\begingroup$ Two people of the same gender are not identical. $\endgroup$ – eyeballfrog Jun 20 '19 at 17:22
  • $\begingroup$ How do you approach the last two problems with non-identical genders? $\endgroup$ – RUNN Jun 20 '19 at 17:30

When dealing with these sort of counting problems, you should try to figure out "what makes one arrangement a 'different' arrangement than another?"

In general, problems about arranging "people" implies that the people are distinct. They each have their own unique name, their own unique life experiences, maybe even their own unique heights or ages. (It is helpful to assume in some cases that their ages are all distinct so that we can have an arbitrary way of ordering them for certain counting arguments).

As such, an arrangement where person $A$ is in one location is considered to be a different arrangement than one where person $B$ is in that location instead, regardless of whether they are the same gender or wearing the same color clothing, etc...

Compare this to a problem about arranging balls where here we instead commonly assume the balls are indistinguishable from one another. Whether it was this red ball in the first position or a different red ball in that position, we can't tell the balls apart so it does not matter.

For your problem, since we are dealing with arranging people, treat each person as being different.

  • How many ways to arrange the eight people in a line for a photograph

There are $8!$ ways. This is seen easily by application of the rule of product and remembering that each person is different. An answer of $\binom{8}{3}$ would have been correct if we couldn't tell people apart by anything except gender.

  • How many ways to arrange the eight people in a line for a photograph if at each end there must be a male?

First pick who the male is on the far left. There are $3$ such choices. Then pick the male who stands on the far right. There are $2$ such choices. Then, from left to right in the remaining available spaces select which person stands there. This gives a final total of $3\times 2\times 6!$ number of arrangements.

An answer of $6$ would have been correct if we couldn't tell people apart in any way except gender.

  • How many ways are there to form a subcommittee if it must contain at least one male?

Here, since we started talking about "committees" and "subcommittees" we now traditionally assume that the order within the committee does not matter (E.g., we don't care who the secretary is, who the leader is, who the subordinates are, etc...).

To count the number of ways to select a subcommittee here where we don't care about there being a male or not, this would simply be $\binom{8}{4}$ number of ways. Since we want at least one male, we can subtract away the "bad" arrangements which are those consisting only of women. There are $\binom{5}{4}$ "bad" arrangements, so the final total number of committees with at least one male is $\binom{8}{4}-\binom{5}{4}$.

As an aside, you could have done this directly rather than indirectly as $\binom{3}{1}\binom{5}{3}+\binom{3}{2}\binom{5}{2}+\binom{3}{3}\binom{5}{1}$ and arrived at the same answer. (A now deleted answer incorrectly claimed your attempt for the third problem was incorrect, despite giving the same answer just written differently)


(i) For the first point, since the question states "In how many different ways can they be arranged for the photo?", I think you just have to consider people, regardless of the gender. Hence it's

$8! = 40320$

(ii) you can choose between 3 males for the beginning of the line, between 2 for the end of the line, and then you sort 6 people in the remaining 6 places.

$3 \times 2 \times 6! = 4320$

(iii) you subcommittee has to contain at least one male, so we can calculate every possible committee and then subtract the ones with only women. The total of possible committee is when we choose 4 people out of 8. The women-only committee are made chosing 4 people out ot 5. Hence

$ 8\choose4 $ $-$ $ 5\choose4 $ $ = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} - 5 = 7 \times 2 \times 5 - 5 = (14 - 1) \times 5 = 13 \times 5 = 65 $

  • 1
    $\begingroup$ You got (iii) incorrect. The answer you gave incorrectly assumes that order within the committee is relevant but traditionally problems phrased in this way order within committees is not relevant. $\endgroup$ – JMoravitz Jun 20 '19 at 17:58
  • $\begingroup$ @JMoravitz thanks, I updated. Your answer is already correct, but I'll leave mine in case somebody wants to read a shorter version of the solution $\endgroup$ – MarMik Jun 20 '19 at 18:10

First Question- Answer for the first question would be 8!. Here we have considered distinctive persons regardless of their gender.

Second Question- We take two men and put them in the corners. What we left is 1 man and 5 women. So for this, the number of arrangements would be 6!. Now remember, we had to choose two me out of three men. So the permutation to choose two men from three men group would be 3!/(3-2)!=3! Answer would be 3!*6!

Third Question- Here we are asked to choose committee of 4 people from 3Men 5 Women group such that committee contain at least 1 men. To the contrary let's suppose this committee does not contain any men. Number of such committees would be 5!/4!*(5-4)!

Number of committees with any people on board is 8!/4!*4!

Therefore the number of committees that contains at least 1 men would be 8!/4!4!-5!/4!(5-4)!

  • $\begingroup$ This tutorial explains how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Jun 20 '19 at 21:08

if everyone is distinct, then
1. Simply permute everyone i.e. $8!$
2.Now for the second part, you have to select two men out of three which is ${3 \choose 2}2!$ to place at ends with permutation, for the arrangement part you have 1 man and 5 women after fixing two men. So, their arrangement is ${3 \choose 2}2!6!$.
3. For the third part,
choose 1 man out of three and 3 women out of 5 i.e $\binom{3}{1}\binom{5}{3}$

choose 2 men out of three and 2 women out of 5 i.e $\binom{3}{2}\binom{5}{2}$

choose 3 men out of three and 1 woman out of 5 i.e. $\binom{3}{3}\binom{5}{1}$
Adding this will give the solution which is $\binom{3}{1}\binom{5}{3}+\binom{3}{2}\binom{5}{2}+\binom{3}{3}\binom{5}{1}$ as posted by JMoravitz.

  • $\begingroup$ In the second question, you forgot to arrange the two men at the ends of the row. In the third question, you are counting each committee with two men twice, once for each way of designating one of the two men as the man on the committee, and each committee with three men three times, once for each way of designating one of the three men as the man on the committee. Please read the answer posted by JMoravitz. $\endgroup$ – N. F. Taussig Jun 20 '19 at 21:12
  • $\begingroup$ Thanks, @N.F.Taussig, I corrected my solution. $\endgroup$ – wild_fox Jun 20 '19 at 21:34

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