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For 3(1), I got (6*3*2*2*1*1)/(6!) for 3(2), my reasoning is that holds the same

ID: 3327939 • Letter: F

Question

For 3(1), I got (6*3*2*2*1*1)/(6!)

for 3(2), my reasoning is that holds the same logic from 3(1), yet combine additional 3 boys. Therefore I got (9*3*2*2*1*1*C(n=6,r=2))/(9!)

Did I make a right approach? If not, could you please enlighten me? Thank you!

3. (1)There are 6 chairs around a table. A group of 3 boys and 3 girls take the seats randomly. What is the probability that no girl sits beside another girl? (2)There are 9 chairs around a table. A group of 6 boys and 3 girls take the seats randomly. What is the probability that no girl sits beside another girl?

Explanation / Answer

Yes you are make right approch, I appriciate your work,

1)

Probability that no girl sits beside another girl =  (6*3*2*2*1*1)/(6!) = 72 /720 = 0.1

2)

Probability that no girl sits beside another girl =  (9*3*2*2*1*1*C(n=6,r=2))/(9!)

= 108 * 15 / 362880

= 0.0045

Probability that no girl sits beside another girl = 0.0045

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