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Neo1
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Hello just a small question from a past paper:

A child's game uses five bricks. One is blue , one is green , one is yellow and two are white. The five bricks are arranged in a line.

(i) How many different arrangements if the colours are there?

(ii) Assuming that all arrangements in part (i) are equally likely, find the probability that the two white bricks are at the ends of the line.


Thanks a lot
From John
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Hmm, this isn't in my textbook, but I think I've worked out how you do it.

Example: How many ways can you arrange "ABB"

= ABB, BAB, BBA, so there are 3 ways.

Total number of letters = 3, total number of letters the same = 2.
3!/2! = 3, the number of ways.

Example: How many ways can you arrange "1233"

= 1233, 1323, 1332, 2133, 2313, 2331, 3123, 3132, 3231, 3213, 3312, 3321 = 12 ways.

Number of numbers = 4, number of numbers the same = 2.
4!/2! = 12, which is the number of ways.

So for your example, there are 5 blocks, and 2 of them are the same. So the number of ways to arrange them is 5!/2! = 60. I tried writing out the different ways and I got 60 as well, so I assume it's right.

If the two white bricks have to be at the end, we're looking to see how many ways we can arange the yellow, green and blue blocks, = 3!

So there are 6 ways to arrange them with the two white blocks at the end, and 60 ways to arrange them in total.

Probability = 6/60

= 1/10.

I hope I'm right, lol
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Chewwy
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5!/2!

p(end of line) = 2/5 * 1/4 = 1/10
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yay
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Neo1
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I put that answer and i got it wrong..i think it means like this:

W ▀ ▀ ▀ W

WW ▀ ▀ ▀

▀ ▀ ▀ W W

so what would the answer be now...3! x 3 / 120 ??


Thanks a lot
From John
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*LEGEND*
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with part b i think the answer is 0.2

Because: there is 2! ways the white bricks can be arranged and 3! ways the coloured bricks can be arranged

So its (2! X 3!) / 60 = 0.2
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