Permutations problem

• Oct 11th 2012, 12:53 AM
miguel11795
Permutations problem
You have three toys, and there are 4 boys and six girls. If you want to give all the toys away, and you want at least two boys to receive a toy, how many ways can this be done.

I'm not sure if I did this correctly, but I took the number of permutations possible $10*9*8$, then I subtracted the number of permutations wherein two boys cannot receive a gift $7*6*5$. This would become: $720-210=510$.

I'm not really sure if what did was right, if my answer is correct, or whether or not there's another way of solving this. Can I ask for help?

• Oct 11th 2012, 02:13 AM
chiro
Re: Permutations problem
Hey miguel11795.

There are many ways to do problems in mathematics so if you see someone doing it one way, just remember that it is not the only way to do a problem.

The way that you did it (the idea as well in general) as a good way. The number of permutations is as you said (720) in total is right.

Now for the number of permutations for no boys getting a gift and one boy getting a gift. We find those and subtract those to get the number of ways for at least two boys getting a gift.

So no boys getting a gift is 6*5*4 = 120, since you only have 6 girls and one boy getting a gift means you can have the boy get a gift either at the first step, the second step, or the third step. So calculate these individually we get:

1st step = 4*6*5 = 120
2nd step = 6*4*5 = 120
3rd step = 6*5*4 = 120

So total number of times 1 boy gets a gift = 360 and number of times boy gets no gift = 120 so these add to 480.

Subtracting 480 from 720 gives us 240 different choices.

So in this problem, we had no boy getting a gift and we also had a boy getting a gift at one of the three stages of getting a gift (and considered all three stages).

This technique also relates to what are called probability trees and they are useful for doing these probabilities. For this problem you have two branches: one where no boy gets a gift and one where one boy gets a gift.

Now the first one with no boy is a simple calculation with all the girls, but the one with one boy is different because each stage can be given to a boy (and the rest must be girls) so we have to look at the three cases individually, so we create a situation where we look at the three things individually and add them up.

In probability if things have nothing in common, then the number of things is added and all three cases are completely separate from one another.

The multiplication rule in probability should be explained so I'll do a short explanation.

Everytime you do a multiplication, the next thing in the multiplication is independent to the things before it. So in your example I had 10*9*8 and with regards to independence I had 10 choices first, then 9 but the 9 choices are completely independent from the 10 I had before and the 8 are independent from the one before that.

So when I start I have 10 choices and then 9 but they are independent choices. If you see rules like probability independence and multiplication, it's important to keep this in the back of your mind.

So as a suggestion, remember to break things up into their smallest parts and then add up all the parts together later on. If you do this, you won't have to rely on rules that are a bit fuzzy and you will also be able to understand your answer in a deeper way.
• Oct 11th 2012, 07:12 AM
Soroban
Re: Permutations problem
Hello, miguel11795!

Quote:

You have three toys, and there are 4 boys and six girls.
You want to give all the toys away, and you want at least two boys to receive a toy.
How many ways can this be done?

I assume that the three toys are distinguishable.

We can give all three toys to the boys.
. . There are: . $_4P_3 = 24$ ways.

We can give two toys to the boys.
. . There are: . $_4P_2 = 12$ ways.
And give one toy to a girl.
. . There are: . $_6P_1 = 6$ ways.
Hence, there are: . $12\cdot6 = 72$ ways.

Therefore, there are: . $24 + 72 \:=\:96$ ways to distribute the toys.
• Oct 11th 2012, 11:25 AM
Plato
Re: Permutations problem
Quote:

Originally Posted by Soroban
And give one toy to a girl.
. . There are: . $_6P_1 = 6$ ways.

There are actually 18 ways to give one toy to one girl.
• Oct 12th 2012, 06:19 AM
peterkelly01
Re: Permutations problem
The way the question is worded does not ptrclude giving a child (a boy) more than one toy!
• Oct 12th 2012, 09:18 AM
johnsomeone
Re: Permutations problem
As peterkelly01 wrote, the technical wording permits a child to receive more than one toy. Is that what's intended?
Also, in order to solve the problem, you need to know if the three toys are considered to be identical or distinct. That's not clear given how the problem is stated.
• Oct 13th 2012, 12:40 PM
johnsomeone
Re: Permutations problem
Here's what I get:

If the toys are indistinguishable and a child can receive more than one toy, then it's:

the number of ways to choose 2 boys times the number of ways to choose any 1 of those 10 children, minus the repetition from things like[{B1, B2}, B4] and, elsewhere in the list, [{B2, B4}, B1].
Can see that, for i not j, [{Bi, Bj}, Bj] has no repetition. But for i<j<k, [{Bi, Bj}, Bk] and [{Bi, Bk}, Bj] and [{Bj, Bk}, Bi] all occur, so for each 1<=i<j<k<=4, this approach counts [{Bi, Bj}, Bk] 3 times when it should've only counted it once.
Thus need to subtract 2 times the number of distinct ways to have 1<=i<j<k<=4, which is 2 times 4.

Num Ways $= \left( \begin{matrix} 4 \\ 2 \end{matrix} \right) \times \left( \begin{matrix} 10 \\ 1 \end{matrix} \right) - 8 = (6)(10) - 8 = 52$.

If the toys are indistinguishable and a child can receive at most one toy, then it's:

the number of ways to choose 2 boys times the number of ways to choose any 1 of those 8 remaining children, minus the repetition from things like [{B1, B2}, B4] and, elsewhere in the list, [{B2, B4}, B1].
Again, for i not j, [{Bi, Bj}, Bj] never appears on the list, so has no repetition. But for i<j<k, [{Bi, Bj}, Bk] and [{Bi, Bk}, Bj] and [{Bj, Bk}, Bi] all occur, so for each 1<=i<j<k<=4, this approach counts [{Bi, Bj}, Bk] 3 times when it should've only counted it once.
Thus need to subtract 2 times the number of distinct ways to have 1<=i<j<k<=4, which is 2 times 4.

Num Ways $= \left( \begin{matrix} 4 \\ 2 \end{matrix} \right) \times \left( \begin{matrix} 8 \\ 1 \end{matrix} \right) - 8 = (6)(8) - 8 = 40$.

If the toys are distinguishable and a child can receive more than one toy, then it's:

All ordered triples (a position identifies a toy received) based on the set {B1, B2, ... B4, G1, G2, ... G6} that contain at least 2 distinct boys. That's all ordered triples less those that have no boys, less those that have exactly 1 boy, less those that exactly 2 boys but who are the same boy, less those that have exactly 3 boys, but who are the same boy.
All ordered triples with no boys counts 6^3.
All ordered triples with exactly one boy = 4 (for the choice of boy) times 3 (for that boy's position) times 6^2 (the ways to populate the rest with girls).
All ordered triples with exactly 2 boys but who are the same boy = 4 (for the choice of boy) times 3 (for that remaining girl's position) times 6 (the ways to populate the rest with girls).
All ordered triples with exactly 3 boys but who are the same boy = 4 (for the choice of boy).

Num Ways $= 10^3 - (6^3) - (4)(3)(6^2) - (4)(3)(6) - (4)$

$= 1000 - 6(6^2) - (12)(6^2) - (2)(6^2) - 4$

$= 1000 - 20(6^2) - 4 = 1000 - 720 - 4 = 276$

If the toys are distinguishable and a child can receive at most one toy, then it's:

All ordered triples (a position identifies a toy received) based on the set {B1, B2, ... B4, G1, G2, ... G6} that contain at least 2 distinct boys, without repetition.
Break that down into all ordered triples of exactly two boys fullfilling that, and all ordered triples of exactly 3 boys fulfilling that.
Count of all ordered triples of exactly two boys fullfilling that = 3 (position of the girl) times 6 (choice of the girl) times 12 (pick one boy, put him in the first open slot, and then pick a remaining boy to put in the last slot).
Count all ordered triples of exactly 3 boys fulfilling that = (4)(3)(2) (pick one boy, then another, then another).

Num Ways $= (3)(6)(12) + (4)(3)(2) = (18)(12) + (2)(12) = (20)(12) = 240$.