Probability Question (1 Viewer)

[fishy89]

This is from the Cambridge 3 unit Year 12 textbook (Exercise 10E Q22):

"In how many ways can a boat crew of eight women be arranged if three of the women can only row on the bow side and two others can only row on the stroke side?"

So far I have:
You need 4 women on the bow side and 4 on the stroke side.
3 of the 8 women are required on the bow side and they can be arranged in 3! ways.
2 of the 8 women are required on the stroke side and they can be arranged in 2! ways.
The remaining 3 women can be arranged in 3! ways.
The group of 3 women and the 4th woman on the bow side can be arranged in 2! ways.
The group of 2 women and the other 2 woman on the stroke side can be arranged in 3! ways.
Total possibilities: 3! x 2! x 3! x 2! x 3! = 864

But the answer is apparently 2 times that: 1,728.
Can somebody please tell me what I'm missing?

 

[iampeterr]

So number of ways of placing particular women on bow side = 4x3x2
Number of ways of placing women on stroke = 4x3
Rest of women = 3!
Total no. = 3! x 4 x 3 x 4 x 3 x 2 = 1728

 

[deswa1]

I do it differently to you. This is my working though:

Call the women B1,B2,B3,S1,S2,W1,W2,W3 (where B means Bow, S means stroke and W means woman)
- B1 can take one of four spots on the bow side, B2 can then take 3 and B2 can take 2. Therefore total combinations so far is 4x3x2
- S1 can take one of four spots on the stroke side, S2 can then take 3. Therefore total combinations here is 4x3 and total combs so far is 4x3x2x4x3
- The remaining three women will take up the remaining three spots in 3! ways. Therefore total combs is 4x3x2x4x3x3!=1728 as required

​Edit: Beaten

 

[fishy89]

Thanks for your help iampeterr and deswa1. I think I'll stick to solving the questions your way rather than trying to solve them in a convoluted way.

 

[deswa1]

Thanks for your help iampeterr and deswa1. I think I'll stick to solving the questions your way rather than trying to solve them in a convoluted way.

Yeah, that's what I'd recommend. I always try solving these questions intuitively rather than with a formula otherwise I stuff them up

 

[fishy89]

I have another question. Q37 from Exercise 10E in the Cambridge 3 unit text.

"Numbers less than 4000 are formed from the digits 1, 3, 5, 8 and 9, without repetition.

d) how many of them are divisible by 3?"

so far i had:
for 3XXX numbers, this includes 3! arrangements with the numbers 951, and 3! arrangements with the numbers 189.
for 1XXX numbers, this includes 3! arrangements with the numbers 359, and 3! arrangements with the numbers 389.
for triple digit numbers, this includes the 3! arrangements for each of the numbers 135, 138, 159 and 189.
for double digit numbers, this includes the 2 arrangements for each of the numbers 15, 18, 39.
for single digit numbers, this includes 3 and 9.
total: 8 x 3! + 2 x 3 + 2 = 56.

The answer at the back is 50 though. What have I included that I should not have?