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#176 Re: Help Me ! » binomial » 2012-12-22 06:17:17

it is wrong.

I think that it would be 4^50.

Every member has 4 choices. 50 members. But im not sure...

#177 Re: Help Me ! » binomial » 2012-12-20 15:33:22

thank you. I have another question:
We have 8 pieces of strawberry candy and 7 pieces of pineapple candy. In how many ways can we distribute this candy to 4 kids?

and:
In how many ways can we distribute 13 pieces of identical candy to 5 kids, if the two youngest kids are twins and insist on receiving an equal number of pieces?

and:

4 students are running for club president in a club with 50 members. How many different vote counts are possible, if members may choose not to vote?

#178 Re: Help Me ! » binomial » 2012-12-20 06:05:25

Another problem:
Consider the polynomial

What are the coefficients of $ f(t-1) $? Enter your answer as an ordered list of four numbers. For example, if your answer were $ f(t-1) = t^3+3t^2-2t+7 $, you'd enter (1,3,-2,7). (This is not the actual answer.)

#179 Re: Help Me ! » binomial » 2012-12-19 10:48:08

so because i have a 2 * 2n-1 choose n, it is bound to be even?

#180 Help Me ! » binomial » 2012-12-19 05:31:42

cooljackiec
Replies: 25

how can we prove 2n choose n  is always even?

#181 Re: Help Me ! » Another random problem » 2012-12-13 13:21:08

actually, bobbym
The point (x,y) satisfies x<y if and only if it belongs to the shaded triangle bounded by the lines x=y,y=2 , and , x=0 the area of which is 2. The rectangle has area 6, so the probability in question is 1/3

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