# combitations

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• July 4th 2007, 12:25 PM
Raiden_11
combitations
1. Two mating fruit flies each have one gene for black eyes and one for red eyes. The offspring receive one eye-colour gene from the male and one from the female.
a) Produce a tree diagram showing all the possible combinations for the offspring.
b) How many distinct combinations can be produced?

2. There are 14 members in a student government. They decide to send a committee of 3 students, one of who will be designated as the spokesperson, to negotiate a new attendance policy with the principal.
a) In how many ways can this be done?
b) Suppose there are 8 women and 6 men in the student government. How many possibilities are there if the committee must have at least 1 person of each gender?

3. How many 8-letter sequences are there consisting of As, Bs, and Cs with the following restrictions?

c) The sequence must have exactly 4 A’s.
d) The sequence must have exactly 3 A’s and 3 B’s.
• July 4th 2007, 12:39 PM
galactus
Why don't you give us some of your thoughts on these. They're not that bad.

How about 2a and 2b?. How many ways are there to choose 3 out of 14?.

If there must be at least one of each gender and 3 are to be chosen, what are the cases?.

3c was answered by plato. For 3d, think about it a little. There are 8 letters, 3 must be A's, 3 must be B's, that means the other two are C's.
• July 4th 2007, 01:26 PM
Raiden_11
well lets see..
2.a) there are 4 ways to choose 3 out of 14.

b)would it be 8x6x3

3d)so the other 2 must be two c's....ok but do you put in in an equation?

and what about #1.....

I'm really sorry, i know it must be frustrating, but i would really appreciate it if you could show me how to do them
• July 4th 2007, 01:50 PM
galactus
I'm not frustrated. I assume this is not homework and you are studying counting problems on your own?.

counting problems normally do not rely on equations so much as deduction and practice.

There are formulas for choosing r objects from n objects when order does not matter: $\frac{n!}{(n-r)!r!}$

When order does matter: $\frac{n!}{(n-r)!}$

2a. There are a lot more than 4 ways to choose 3 out of 14.

$\frac{14!}{(14-3)!3!}$

2b. There are 8 women and 6 men. You are to choose 3, but there must be at least 1 of each. There must either be 2 men and 1 woman or 2 women and 1 man. $C(6,2)C(8,1)+C(8,2)C(6,1)$

3d. Since there is an 8 letter sequence consisting of 3 A's 3 B's, then the other 2 letters must be C's.

$\frac{8!}{3!3!2!}$
• July 4th 2007, 05:19 PM
Raiden_11
Just one more?
Thankyou so much,if you can just show me how to do #1a and b.....thats all......thankyou again!!!!