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Permutations and Combinations (1 Viewer)

Sparklingstar

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Hi :jump:,
Can someone please help me with this question?

A girl has 6 pairs of earrings, 3 t-shirts, 2 pairs of jeans and 8 pairs of shoes. If she wishes to choose one of each article, how many ways can she dress up to go out?
The answer is 288

Thankyou :)
 

enoilgam

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*Moved to Maths Extension 1 Forum*

This may sound stupid (I dont know if it is correct), but wouldnt you do 3x2x8x6 = 288, because there are no restrictions on the combinations. I'm not sure at all though.
 

Carrotsticks

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This may sound stupid (I dont know if it is correct), but wouldnt you do 3x2x8x6 = 288, because there are no restrictions on the combinations. I'm not sure at all though.
Still got it in you I see :p
 

Shadowdude

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This may sound stupid (I dont know if it is correct), but wouldnt you do 3x2x8x6 = 288, because there are no restrictions on the combinations. I'm not sure at all though.
That's right.


A girl has 6 pairs of earrings, 3 t-shirts, 2 pairs of jeans and 8 pairs of shoes. If she wishes to choose one of each article, how many ways can she dress up to go out?

1. Choose a pair of earrings... 6 ways
2. Choose one T-shirt... 3 ways
3. Choose a pair of jeans... 2 ways
4. Choose one pair of shoes... 8 ways

Answer: 6*3*2*8 = 288


Yet again, pro-tip: set out your working clearly.
 

Sparklingstar

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How about this question?

A family consists of a father, a mother, 3 girls and 4 boys.
If the same family is seated randomly at a round table, find the probability that the parents are separated by exactly two seats, and these seats are both occupied by boys?

Thanks:)
 
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Shadowdude

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How about this question?

A family consists of a father, a mother, 3 girls and 4 boys.
If the same family is seated randomly at a round table, find the probability that the parents are separated by exactly two seats, and these seats are both occupied by boys?

Thanks:)
This is a 'successes' / total possibilities probability question.

Total possibilities to seat nine people on the round table: 8!

A 'success' is when the parents are separated by two seats AND these seats in between are both occupied by boys

1. Seat a parent at the table... 2 ways
2. Choose 2 boys to seat next to the first parent... C(4,2) ways
3. Arrange these boys in the two seats... 2! ways
4. Seat the other parent next to these two boys... 1 way
5. We have three girls and two boys remaining, arrange them in the remaining 5 chairs... 5! ways

Answer: 2*C(4,2)*2!*1*5! / 8!

= 1/14.


In the words of Rainbow Dash: OHHHHHHHH YEEAH
 

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