Permutations - arrangement of a boat crew of 8 women

AI Thread Summary
The discussion focuses on calculating the number of arrangements for a boat crew of 8 women with specific constraints: 3 women can only row on the bow side and 2 can only row on the stroke side. The initial calculation suggested using permutations without considering the constraints, leading to a total of 8! arrangements. However, the correct approach involves recognizing the fixed positions for the constrained rowers and permuting the remaining women accordingly. The final answer, after accounting for these constraints, is determined to be 1728 arrangements. The discussion emphasizes the importance of understanding the seating limitations and the order of individuals in the arrangement.
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Homework Statement


How many ways can a boat crew of 8 women be arranged if 3 of the women can only row on the bow side and 2 can only row on the stroke side?

Homework Equations


The Attempt at a Solution


I simply did 8!/5! (5!/3!)3! which is simply 3!(8!/3!)

The correct answer is 1728.
 
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And 3!(8!/3!) is simply 8!, which is the total number of ways you can arrange the crew without taking the constraints into account.

Let's try to get some intuition first... let B denote someone who can only row on the bow side, S on the stroke side, and A someone who can be on either side.
One possible arrangement is:

A A
B A
B S
B S

You need to realize there are two possible ways of permuting this
(1) you can move the A's around on the left hand side or the right hand side, e.g.
B A
A S
B A
B S

So you will have to figure out the number of permutations this gives you.

(2) in addition, all the rowers are individuals whose order matters. So for any variation you get from the above, you can move the B's, the S's and the A's, e.g. if we number the ladies then the first layout above can either correspond to
A1 A2
B1 A3
B2 S1
B3 S2

or to

A2 A3
B2 A1
B3 S2
B1 S1

or any other combination which again leads to (A, B, B, B; A, A, S, S) when you remove the numbers.
 
You can try this also:

The four seats on the bow side will "select three seats" in which the the three women will sit. The order does not count. Similar for the stroke side, which will "select two seats", and the order does not count. After women are seated on each of the sides, there are three women to be seated in 3 places in any order.
 
You can also reduce the full permutation successively by eliminating the proportion of "wrong" permutations.

So first considering the bow-siders, the first could have been in 8 seats of which 4 are correct (multiply by 4/8); the second could have been in 7 (remaining) seats of which 3 are correct, then 2 from 6 seats, and then on the stroke-siders, 4 from 5 seats, and 3 from 4 seats.
 
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