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# Help needed with some S1 questions watch

1. Each of 6 cards has a different single letter written on it. The letters on the cards are A, B, C, D, E and F. The cards are shuffled and then placed in a row.
(i) How many different possible arrangements of letters are there? [1]
(ii) In how many of these arrangements are the vowels next to each other? [4]

Only need help on part (ii) for this, the answer is 5! X 2! = 240.

And then another one: A child’s game uses 5 bricks. One is green, one is blue, one is yellow and two are white. The five bricks are arranged in a line.
(i) How many different possible arrangements of the colours are there?
(ii)Assuming all arrangements are equally likely, what is the probability that the two white bricks are at the ends of the line?

Again it is only part (ii) I need help with: the answer is 3x3!/60=0.3. I got this answer but I am unsure of whether or not the arrangements I put for the white bricks are correct.

Any help would be greatly appreciated.
2. (Original post by Azizz)
Each of 6 cards has a different single letter written on it. The letters on the cards are A, B, C, D, E and F. The cards are shuffled and then placed in a row.
(i) How many different possible arrangements of letters are there? [1]
(ii) In how many of these arrangements are the vowels next to each other? [4]

Only need help on part (ii) for this, the answer is 5! X 2! = 240.

And then another one: A child’s game uses 5 bricks. One is green, one is blue, one is yellow and two are white. The five bricks are arranged in a line.
(i) How many different possible arrangements of the colours are there?
(ii)Assuming all arrangements are equally likely, what is the probability that the two white bricks are at the ends of the line?

Again it is only part (ii) I need help with: the answer is 3x3!/60=0.3. I got this answer but I am unsure of whether or not the arrangements I put for the white bricks are correct.

Any help would be greatly appreciated.
1. (ii) Think about the number of vowels and the number of non-vowels (2 and 5 respectively). If the vowels must be together, there are 2! ways of arranging them. For each of these ways, there are 5! ways to arrange the non-vowels. And voila! 240 arrangements.

For 2. (ii) I didn't get what you got for this one, but I could be wrong:

The two whites can be either at the start or the end of the line.

At the start: (2! x 3!)/2! = 3! [arrangements of whites multiplied by arrangements of green, blue and yellow, and then divided by the whites again to account for them being the same colour]

At the end: (3! x 2!)/2! = 3! [same reasoning, but reversed]

So the probability is (3! + 3!)/60 = (2 x 3!)/60 = 12/60 = 0.2

I think:/ Perhaps someone else can have a go?.....

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Updated: January 23, 2015
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