Question 849370
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Those answers above are all wrong.
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1) in how many ways can 10 fruits be split into 3 parts consisting of 4,2,4 fruits respectively.
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Suppose the first part consisting of 4 is labeled part A
Suppose the second part consisting of 2 is labeled part B
Suppose the third part consisting of 4 is labeled part C

If it doesn't matter what fruits are in what part then the answer
would be C(20,4)C(16,2)C(12,4) but since part A and part C are 
indistinguishable, we must divide by 2! or 2.

[C(20,4)C(16,2)C(12,4)]/2! = (4845*120*495)/2 = 287793000/2 = 143896500
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2)in how many ways can a committee of 5 Doctors be chosen from 9 Doctors. 
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That's just C(9,5) = 126.  (The other tutor set it up right, but worked it
wrong and got twice that.) 
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3) A basket contains 8 Oranges, 3 Mangoes and 9 Apples. if 3 fruits are drawn at random, determine the probability that 1 of each fruit is drawn.
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We find the probability of the complement event.  That's the number of ways to
fail to get 1 of each fruit.

First we find the number of ways to fail to get 1 of each fruit = 
The number of ways to get all three the same fruit or 2 of 1 and 1 of another)

P(3 oranges)+P(3 mangos)+P(3 Apples) + 
P(2 oranges and 1 mango) + P(2 oranges and 1 apple) + P(2 mangoes and 1 apple) +
 
P(1 oranges and 2 mango) + P(1 oranges and 2 apples) + P(1 mangoes and 2 apples)

C(8,3)+C(3,3)+C(9,3)+

C(8,2)C(3,1)+C(8,2)C(9,1)+C(3,2)C(9,1)+

C(8,1)C(3,2)+C(8,1)C(9,2)+C(3,1)C(9,2) =

56+1+84 + 84+252+27 + 24+288+108 = 924

The number of ways to select any 3 fruits from the 20 is C(20,3)

P(fail to get 1 of each) = 924/1140 = 77/95 = 0.8105263158

P(succeeding by getting at least 1 of each) = 1-0.8105263158 = 0.1894736842 

Edwin</pre>