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u-borat

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there are six alphas, 5 betas and 4 gammas lined up in a line.

i)how many arrangements are there where the first alpha PRECEDES the first beta?

ii) how many arrangements are there where the first alpha PRECEDES the first beta which PRECEDES the first gamma?
 

Affinity

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i)
method one:
There are 15! arrangements (imagine the letters are all different first)... if you assign equal probability to these.. you will see that the probably that an alpha comes before a beta is 6/11
so you get 15! * 6 / 11
then divide by (4!5!6!) to mod out the permutations to get

15! * 6 / [11(4!5!6!)]

method two: use some brute force counting.. fixing the first alpha.. the onl gammas can precede it.

ii.
using a similar argument, the probability of the first alpha preceding the first beta AND the first gamma is 6/15. conditioned on that the probability of the first beta preceding the first gamma is 5/9 so

15! * 6 * 5 / (15* 9* 4!5!6!)

alternatively, the first letter must be alpha,

14! * 5/ (9*4!*5!*5!)
 

u-borat

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nice solution; i'm not sure how you can say that the probability than an alpha comes before a beta is 6/11 and co...can u explain that bit?
 

samwell

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Affinity said:
i)
method one:
There are 15! arrangements (imagine the letters are all different first)... if you assign equal probability to these.. you will see that the probably that an alpha comes before a beta is 6/11
so you get 15! * 6 / 11
then divide by (4!5!6!) to mod out the permutations to get

15! * 6 / [11(4!5!6!)]

method two: use some brute force counting.. fixing the first alpha.. the onl gammas can precede it.

ii.
using a similar argument, the probability of the first alpha preceding the first beta AND the first gamma is 6/15. conditioned on that the probability of the first beta preceding the first gamma is 5/9 so

15! * 6 * 5 / (15* 9* 4!5!6!)

alternatively, the first letter must be alpha,

14! * 5/ (9*4!*5!*5!)
cool method thanks.
 

Affinity

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u-borat said:
nice solution; i'm not sure how you can say that the probability than an alpha comes before a beta is 6/11 and co...can u explain that bit?
there are 11 alphas and betas all together. the probability of a particular one coming first is 1/11 and there are 6 alphas. so 6/11
 
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