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# ​ How do I solve this probability question?

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I need b and c, thanks in advance.

Jun 19, 2022

#1
+118617
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But the answes are already there, arn't they?

Jun 20, 2022
#2
+33616
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For

a) How many ways are there of choosing 4 from 14

b) How many ways of choosing 4 from 14 less how many ways of choosing 2 from 12

c) How many ways of choosing 4 from 12 plus how many ways of choosing 2 from 12.

(In general:  How many ways of choosing m from n is given by:  $$\frac{n!}{(n-m)!\times m!}$$ )

Jun 20, 2022
#3
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is it possible to explain b and c. I don't get why "choosing 4 from 12 " and "choosing 2 from 12"?

Guest Jun 20, 2022
#4
+118617
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Lets see

a)  4 people  are chosen from 14

That is just 14 choose 4 = 14C4  = 1001     (it is not really choose, C stands from combinations)

You say you get that?

b)   2 of the candiates do not want to go together.

ok  Let's say those 2 are BOTH chosen

Then there will be 2 chosen out of the remaining 12  which is    12C2  = 66

BUT those are the combinations that are NOT allowed so the number of possible combinations will be

14C4 - 12C2   = 1001 - 66 = 935

Does that help?

Jun 20, 2022
#5
+1

why is it out of 12 and not 14. like C(14,2)

Guest Jun 20, 2022
#6
+118617
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Becasue the two that are not allowed to go together have been taken out 14-2 = 12

What Alan and I have said is

"Lets see how many of the 1001 combinations CONTAIN both those 2 people.  "

So now we have already chosen those two candidates.

We have 12 candidates left and we have to choose 2 of them  12C2

That is how many possible delegations of 4 that contain the 2 people who do NOT want to be together

That is the number of possible delegations that must be NOT counted

The number left AFTER these ones are NOT counted are removed      14C4  -  12C2

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Lets's look at an easier questions,

We'll chose 2 letters from ABC

3 letters 2 to chose that is 3C2=3 ways        AB, AC, BC

BUT      A abd B cannot both be chosen

How many of the 3 ways conatin both A and B

there is one left to choose from and we need to chose 1 so there must be 1 combination with both A and B

(3-2)C(3-2) = 1C1 = 1

All the other combinations do NOT conatin both A and B

what is left is         3C2- 1C1 =  3-1  = 2

Melody  Jun 20, 2022