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Combinations-Number of choices

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1. Joey is having a party. He has 10 friends, but his mom told him he could only invite 6 of them. How many choices are there if
a. there are no restrictions
b. there are 2 brothers who will only attend if they can attend together
c. there are 2 girls who each will not attend if the other one does

https://brainmass.com/math/combinatorics/combinations-number-choices-32738

SOLUTION This solution is FREE courtesy of BrainMass!

a) there are no restrictions
Total no friends= 10
If there are no restrictions then the no of choices = no of ways of choosing 6 out of 10 =
10 C 6 = (10 x 9 x 8 x 7 x 6 x 5 ) / ( 1 x 2 x 3 x 4 x 5 x 6 ) = 210

b ) there are 2 brothers who will only attend if they can attend together

There are 2 cases here:

Case 1 ) Both brothers attend
In this case there are 4 more friends to invite from the remaining 10-2 = 8 friends

no of ways of choosing 4 out of 8 =
8 C 4 = (8 x 7 x 6 x 5 ) / ( 1 x 2 x 3 x 4) = 70

Case 2 ) Both brothers do not attend
In this case there are 6 friends to invite from the remaining 10-2 = 8 friends

no of ways of choosing 6 out of 8 =
8 C 6 = (8 x 7 x 6 x 5 x 4 x 3 ) / ( 1 x 2 x 3 x 4x5x6) = 28
Therefore total no of choices= 70 + 28 = 98

c) there are 2 girls who each will not attend if the other one does

We can calculate this problem by first calculating the no of ways in which the two girls are together.
We will subtract it from total no of ways of choosing where there are no restrictions to get the no of ways in which the two girls are not together.

If there are no restrictions then the no of choices = no of ways of choosing 6 out of 10 =
10 C 6 = (10 x 9 x 8 x 7 x 6 x 5 ) / ( 1 x 2 x 3 x 4 x 5 x 6 ) = 210

No of cases in which both girls attend
In this case there are 4 more friends to invite from the remaining 10-2 = 8 friends

no of ways of choosing 4 out of 8 =
8 C 4 = (8 x 7 x 6 x 5 ) / ( 1 x 2 x 3 x 4) = 70

Therefore total no of ways in which the friends can be invited so that they are not together = 210-70= 140