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2. The binomial theorem gives us (2x-1)^5=5C0(2x)^5*(-1)^0+5C1(2x)^4*(-1)^1+5C2(2x)^3(-1)^2+5C3+(2x)^2(-1)^3+5C4(2x)^1(-1)^4+5C5(2x)^0(-1)^5=\(32x^5-80x^4+80x^3-40x^2+10x-1\), so \(\boxed{-40}.\)

3. Try to use the binomial expansion.

4. How many men and woman can you choose?

5. Find the total number of paths to A-C, then from C-B.

1. Since the order you pick the people in for the groups does not matter, you can use combinations to solve this problem. There are \(\binom{9}{4}=126\) ways to pick the people in the group of four. For the next group there are 5 people left that are not in a group so there are \(\binom{5}{3}=10\) ways to pick the group of three people. The remaining two people will be the group of two. The total number of ways to pick the groups is \(126 \cdot 10 = \boxed{1260}\)

5.Find the number of paths from  to  that pass through point  (Each step can only go up or to the right.)

Note that a set of moves that get us from  A to C  is  just  (R, R, U, U, U)

And we can choose any 2 of 5 places for R in the set and still get from A to C

So.....the total number of paths from A to C  is  C(5,2)   =  10

Likewise

A set of moves that get us from  C to B  is just (U, R,R,R )

And we can choose any 1 of the 4 places for U in the set and still get from C to B

So....the the total  paths from C to B is just  C(4,1)  = 4

And the total possible paths   =  C(5, 2) * C(4, 1)    =   10 * 4   =    40

3. Expand (a+b)^3

We only need  the 3rd row of Pascal's Triangle for the coefficients  [ note that the first row  = Row 0 ]

So....we have

1a^3  + 3a^2b  + 3ab^2  + 1b^3  =

a^3  + 3a^2b + 3ab^2 + b^3

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