Question 1205259: In a cars carrier, how many ways are there to put 8 cars labeled as {A, B, ..., H} in 4 containers labeled as { 1, 2, 3, 4 }? Each container must include at least one car.
Your answer is
40824
If the containe #1 must have the car A in it, find the new number of possible ways.
Your answer is
5040
are my answers correct?
Found 2 solutions by ikleyn, Edwin McCravy:
Answer by ikleyn(52921)   (Show Source):
You can put this solution on YOUR website! .
For the first question, it is not YOUR answer.
It is MY answer, which I provided to you in my post
https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1205233.html
Call the things by their proper names.
Answer by Edwin McCravy(20066)  (Show Source):
You can put this solution on YOUR website!
I'll just do the first part. There are just these 5 ways to
break up 8 cars into a sum of 4 positive integers, without
regard for order:
1. 1+1+1+5 = 8
2. 1+1+2+4 = 8
3. 1+1+3+3 = 8
4. 1+2+2+3 = 8
5. 2+2+2+2 = 8
Of course we must choose and order each of the 5 cases.
Case 1: 5 cars in 1 container, 1 car in each of the other 3.
We can choose the container to hold 5 cars in 4 ways.
We can choose the 5 cars for that container C(8,5)=56 ways. The other
3 cars can be distributed to the other 3 containers in 3!=6 ways.
That's (4)(56)(6) = 1344 ways
Case 2: 4 cars in 1 container, 2 cars in another container, and 1 car in
each of the other 2 containers.
We can choose the container to hold 4 cars in 4 ways.
We can choose the 4 cars for that container C(8,4)=70 ways.
We can choose the container to hold 2 cars in 3 ways.
We can choose the 2 cars for that container in C(4,2)=6 ways. The other
2 cars can be distributed to the other 2 containers in 2!=2 ways.
That's (4)(70)(6)(2) = 3360 ways
Case 3: 3 cars each in 2 containers, and 1 car in each of the other 2 containers.
We can choose the 2 containers to hold 3 cars in C(4,2)=6 ways.
We can choose the 3 cars for the container with the smaller number C(8,3)=56 ways.
We can choose the cars for the container with the larger number C(5,3)=10 ways.
The remaining 2 cars can be distributed to the other 2 containers in 2!=2 ways.
That's (6)(56)(10)(2) = 6720 ways
Case 4: 2 cars each in 2 containers, 3 cars in 1 container, and 1 car in 1 container
We can choose the 2 containers to hold 2 cars in C(4,2)=6 ways.
We can choose the 2 cars for the container with the smaller number C(8,2)=28 ways.
We can choose the 2 cars for the container with the larger number C(6,2)=15 ways.
We can choose the container to hold 3 cars in C(2,1)=2 ways.
We can choose the 3 cars for that container in C(4,3)=4 ways.
The container to hold only 1 car gets the 1 car left in 1 way.
That's (6)(28)(15)(2)(4)(1)=20160 ways
Case 5: Each container holds 2 cars each.
We can choose the 2 cars for the container with the number 1 in C(8,2)=28
ways.
We can choose the 2 cars for the container with the number 2 in C(6,2)=15 ways.
We can choose the 2 cars for the container with the number 3 in C(4,2)=6 ways.
We can choose the 2 cars for the container with the number 4 in C(2,2)=1 way.
That's (28)(15)(6)(1)=2520
The total = 1344+3360+6720+20160+2520 = 34104
I haven't gone over this with a fine-tooth comb to see if I've made an error.
I'll check it later, and correct it, or maybe another tutor will solve it.
Edwin
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