Permutations exercises with alike objects; check my homework

could you please review my answers and confirm my reasoning. This is self-teaching work.

In how many ways can 8 people be seated in a row if:

(a) there are no restrictions:

(b) persons A and B must sit next to each other:

Answer: 8!/2! Now, the demoninator is 2! because the couple are considered 'a like', right?

(c) there are 4 men and 4 women and no 2 men and 2 women can sit next to each other

Answer: 8!/(4!*4!) I think this is correct, can you explain why?

(d) there are 5 men and they must sit next to each other.

Answer: 8!/5! I can't explain why this is correct

(e) there are 4 married couples and each couple must sit together

8!/(2!X4) the 4 couples are alike.

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Let me discuss these answers stepwise. These problems should be thought in a two step process. You can easily realize that two humans can not be treated alike as can be the colors (red, orange etc.). Different person has different identity. So, most of your answers are not correct as you have taken persons to be alike. Don't worry, this is a very common mistake that almost everyone makes initially.

(A) As you have rightly pointed out, when there are no restrictions, the answer for (A) is 8!.

(B) If you want to place $n$ things in $n$ positions, and you care about order, you can do it in $n!$ ways. (Doing so is called permuting $n$ items at a time.)

So, your reasoning is not proper for (B). Here persons A and B can be thought of as a block of items along with 6 other individual items (here persons). So, as a whole we get 7 items (1 block+6 items). They can be arranged in $7!$ ways. You can then permute the 2 items (person A and B here) within the block (or within themselves) in $2!$ ways. So, the answer should be $7!\times2!$.

(C) To preserve this condition you should either select seat 1, 3, 5, 7 for men and seat 2, 4, 6, 8 for women or seat 2, 4, 6, 8 for men and seat 1, 3, 5, 7 for women. Then you should permute men and women between them. So, the result should be $2\times4!\times4!$.

(D) This one is very similar to (B). Here 5 men form a block. Along with the other 3 persons they form a total of 4 items. They can be arranged in $4!$ ways. Again these 5 men can be arranged within themselves in $5!$ ways. So, the answer should be $4!\times5!$.

(E) If we arbitrarily name the people A1, A2, B1, B2, C1, C2, D1, and D2, then one possible seating arrangement is B2, B1, A1, A2, C2, C1, D1, D2. Another such assignment might be B1, B2, A2, A1, C1, C2, D1, D2. These two examples illustrate that counting here is a two-step process. First, we have to figure out how many ways the couples A, B, C, and D can be arranged. Permuting 4 items (couples) 4 at a time... there are $4!$ ways. Then, we have to arrange the partners within each couple. Well, couple A can be arranged $2!$ ways. We can even list the ways... they can sit either as A1, A2 or as A2, A1. Likewise, couple B can be arranged in $2!$ ways, as can couples C and D. The Multiplication Principle then tells us to multiply all the numbers together. Four married couples can be seated in a row of 8 seats in:

$4!\times2!\times2!\times2!\times2! = 384$ ways

if each married couple is seated together.

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Thanks Blain Waan. You are accurate with my foundational issue. Early on I recognized that grouping objects (people) with similar attributes is not the same as grouping identical objects (flags). You have provided to me some useful techniques, and I am grateful. –  1c1cle Oct 23 '12 at 19:09
You are most welcome. –  Blain Waan Oct 23 '12 at 19:10