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