A tennis coach divides her 9-player squad into three 3-player groups with each player in only one group. How many different sets of three groups can be made?
I have trouble with these too but I do believe you have doubled counted.
I am not sure what the answer is but maybe it is just what you have divided by 3! because the order the groups are selected in is of no consequence.
That would bring your answer down to 280.
I did this a different way and came up with 280 then too.
Do you know what the correct answer is ?
The first derivative gives the gradient of the tangent.
At stationary points the gradient must be 0.
So find the first derivative (differentiate it) and put that equal to 0.
Solve this equaton and you will have x value.
Sub that into the initial f(x) equation to get the y value. In this case y is called f(x).
Do as much as you can of that and then I can help more.
Means 1 person got 5 and the other 2 people got 1.
There are 3 ways that this can happen becasue the 5 can go to person A,B or C.
then the other two get the same number each.
Means 1 person got 4, one got 2 and 1 got only one.
The 4 can go to person AB or C that is 3 possible ways, then there are two people left. Either of those can get the single card, so that is 2 ways, the last person gets the others. So that is 3*2=6 ways.
So far I have not paid attention to the fact that the cards are different from each other. I dealt with that afterwards.
Try digesting that and then if you need me to I can continue to explain the next bit.
Mack the bug starts at (0,0) at noon and each minute moves one unit right or one unit up. He is trying to get to the point (5,7). However, at (4,3) there is a spider that will eat him if he goes through that point. In how many ways can Mack reach (5,7)?
I get 617 ways. Can you please confirm if you think this is correct or not.
If you believe you know the answer then please state what it is.
Here is an outline of my logic.
There are 35 ways to get from (0,0) to (4,3)
There are 5 ways to get from (4,3) to (5,7)
So there are 35*3=175 ways to get from (0,0) to (5,7) passing through (4,3)
There are 792 ways to get from (0,0) to (5,7)
So there must be 792-175=617 ways to get from (0,0) to (5,7) without passing through (4,3)
My logic seems ok to me.
How many ways are there to form a 7-digit code where each digit can be from 0 to 9 and the product of all the digits in the code is 10,000?
Thanks Geno and Mathmathj28,
Since there is disagreement I will add my 10cents worth.
factor(10000) = 2^4*5^4
So I can only use 2,4,8,5,1 and 5 must be used 4 times
10000/5^4 = 16
5 5 5 5
I need 3 digits that multiply to 16 8,2,1 4,4,1 4,2,2, and that is it.
So the digits can be
5 5 5 5 8 2 1 7!/4! =210
5 5 5 5 4 4 1 7!/(4!2!) = 105
5 5 5 5 4 2 2 7!/(4!2!) = 105
210+105+105 = 420 ways
I just redid part B my own way and it agrees with CPhil's answer ... but his is much easier.
Anyway, here is my logic
(b) I also bought 7 different postcards. How many ways can I send the postcards to my 3 friends, so that each friend gets at least one postcard?
The original answer was 15, that was if all the cards were the same.
|possible numbers of bags each (total 7)|
|5||1||1||3 ways if all the same||7*6=42||3*42=126|
|4||2||1||6 ways if all the same||7*6C2=105||6*105=630|
|3||3||1||3 ways if all the same||7*6C3=140||3*140=420|
|3||2||2||3 ways if all the same||7C2*5C2=210||3*210=630|
|Total||15 ways if all the same||1806 ways if all are different|
So my answer agrees with CPhill's 1806 ways