We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
-1
208
5
avatar+128 

1. A group of nine people is be split into three teams, so that one team has four people, one team has three people, and one team has two people. How many ways can this be done?

 

2. Find the coefficient of x^2 in the expansion of (2x-1)^5

 

3. Expand (a+b)^3

 

4. A club with five men and six women wish to form a committee. The number of men must be between 1 and 3 (inclusive), and the number of women must be between 2 and 4 (inclusive). How many different committees can be formed?

 

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

https://latex.artofproblemsolving.com/6/8/0/680f29b2cbeaa794df57bedf2998ca3aef02aeea.png

 May 19, 2019
 #1
avatar+4330 
+1

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.

 May 19, 2019
 #2
avatar+196 
0

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 \(126 \cdot 10 = \boxed{1260}\)

.
 May 19, 2019
 #3
avatar+105277 
+1

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

 

 

cool cool cool

 May 19, 2019
 #4
avatar+105277 
+1

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

 

 

cool cool cool

 May 19, 2019
 #5
avatar+105935 
+1

Please put different questions on different post!

 May 20, 2019

12 Online Users