1. combination

Hi, I am new so if I do something wrong pls correct me. I have a problem to solve and I am a little confused. Here it is:

We have 10 red balls, 8 green balls and 6 blue balls. I must calculate the ways that all the balls can be shared to 3 discrete boys (without minding the order that the balls are given) so that each kid takes at least one ball of each colour.

After searching and studying I came to this conclusion:

with n=24 balls and k=3 boys :

(n+k-1)!/n!(k-1)!=325 but I am not so sure.  thank you a lot.

2. Originally Posted by milagros Hi, I am new so if I do something wrong pls correct me. I have a problem to solve and I am a little confused. Here it is:

We have 10 red balls, 8 green balls and 6 blue balls. I must calculate the ways that all the balls can be shared to 3 discrete boys (without minding the order that the balls are given) so that each kid takes at least one ball of each colour.

After searching and studying I came to this conclusion:

with n=24 balls and k=3 boys :

(n+k-1)!/n!(k-1)!=325 but I am not so sure.  thank you a lot.
Nope. Looks incorrect to me.

I would have done like this
First give each kid - one ball of each color. There is only one way to do that.
After this we are left with 7R, 5G, 3B balls., correct?
Now we are free to distribute them in any way we like. Let's start with one color.

7R - can be distributed in 9!/7!2! ways = 36
5G - 7!/5!2! = 21 ways
3R - in 10 ways

Final answer - 7560 ways !

3. Originally Posted by aman_cc Nope. Looks incorrect to me.

I would have done like this
First give each kid - one ball of each color. There is only one way to do that.
After this we are left with 7R, 5G, 3B balls., correct?
Now we are free to distribute them in any way we like. Let's start with one color.

7R - can be distributed in 9!/7!2! ways = 36
5G - 7!/5!2! = 21 ways
3R - in 10 ways

Final answer - 7560 ways !
You ' re right . Just found an example . At least I got the formula right ((n+k-1)!/n!(k-1)!). That's a big step for my first attempt with discrete maths, correct ?  Thank you a lot.

4. A drawer contains 8 distinct pairs of gloves each of which consists of a matching left and right glove. If you pull out four individual gloves at random, what is the probability that you have at least one pair of matching gloves? What is the probability you have two pairs of matching gloves?

SOLUTION

There are (16 4) ways to select 4 of the 16 individual gloves. There are 16*14*12*10/4! ways to choose 4 non-matching gloves, so the probability of getting at least one pair of matching gloves is 1  ((16*14*12*10)/4!) / (16 4) = 1  8/13 = 5/13. There are (8 2) ways to choose two of the eight matching pairs, so the probability of getting two matching pairs is (8 2) / (16 4) = 1/65.

Sorry I found this problem but I dont understand what formula he does use for every result he finds. Can you help me? Thanks again.

5. Originally Posted by milagros A drawer contains 8 distinct pairs of gloves each of which consists of a matching left and right glove. If you pull out four individual gloves at random, what is the probability that you have at least one pair of matching gloves? What is the probability you have two pairs of matching gloves?

SOLUTION

There are (16 4) ways to select 4 of the 16 individual gloves. There are 16*14*12*10/4! ways to choose 4 non-matching gloves, so the probability of getting at least one pair of matching gloves is 1 – ((16*14*12*10)/4!) / (16 4) = 1 – 8/13 = 5/13. There are (8 2) ways to choose two of the eight matching pairs, so the probability of getting two matching pairs is (8 2) / (16 4) = 1/65.

Sorry I found this problem but I dont understand what formula he does use for every result he finds. Can you help me? Thanks again.

He is not using any formula. These are just combinatorial arguments. Take each bit one at a time and work it out. Post any specific question you have.

6. Originally Posted by aman_cc He is not using any formula. These are just combinatorial arguments. Take each bit one at a time and work it out. Post any specific question you have.
You re right. Didn't put it correct. First : How does he analyse
(16 4)=16*14*12*10/4!

It must be some type right? Hope you understand what I mean. Thank you.

7. Originally Posted by milagros You re right. Didn't put it correct. First : How does he analyse
(16 4)=16*14*12*10/4!

It must be some type right? Hope you understand what I mean. Thank you.
Like he said this is number of ways to select so that there is no pair. Do you understand this? If not - please sit down and work it out. What argument would you use? Give it some time and be patient.

There is no formula here.

8. Originally Posted by aman_cc Like he said this is number of ways to select so that there is no pair. Do you understand this? If not - please sit down and work it out. What argument would you use? Give it some time and be patient.

There is no formula here.
I will thank you. But given the type c(n,k)=n!/k!(n-k)! I find another solution. But I will work it out. Thank you 9. Originally Posted by milagros I will thank you. But given the type c(n,k)=n!/k!(n-k)! I find another solution. But I will work it out. Thank you If you are attempting these problems, it would be assumed you know well about c(n,k)=n!/k!(n-k)! - How/why do you get this?
If not, its a good time to go back to your notes and wotj these out yourself.

This problem uses no more concepts that that. Most of the time the problems of combinotrics are tough because we need to break it up into simpler problem (which we can count easily), which comes with practice.

So try it out. Post your argument - how would you approach the problem "Select 4 gloves - such that there is no pair"

10. Originally Posted by aman_cc If you are attempting these problems, it would be assumed you know well about c(n,k)=n!/k!(n-k)! - How/why do you get this?
If not, its a good time to go back to your notes and wotj these out yourself.

This problem uses no more concepts that that. Most of the time the problems of combinotrics are tough because we need to break it up into simpler problem (which we can count easily), which comes with practice.

So try it out. Post your argument - how would you approach the problem "Select 4 gloves - such that there is no pair"
Ok just be patient. Just started studying. Its hard for me.  11. Still cant get it. If I have :

A drawer contains 10 distinct pairs of shoes each of which consists of a matching left and right. What are the probabilities in 8 selections not to choose a pair ?

Helppppppppppppppp.

12. Originally Posted by milagros Still cant get it. If I have :

A drawer contains 10 distinct pairs of shoes each of which consists of a matching left and right. What are the probabilities in 8 selections not to choose a pair ?

Helppppppppppppppp.
Let's work it out
So we need to pick 8 shoes - no two of which should be of same pair. Correct?

So each shoe is from a different pair. So selecting 8 shoes is like
1. First select 8 pairs (out of 10)
2. From each of the 8 pairs selected above, select 1 shoe (out of 2)

How many ways?
10C8*2^8
Can you complete?

13. Originally Posted by aman_cc Let's work it out
So we need to pick 8 shoes - no two of which should be of same pair. Correct?

So each shoe is from a different pair. So selecting 8 shoes is like
1. First select 8 pairs (out of 10)
2. From each of the 8 pairs selected above, select 1 shoe (out of 2)

How many ways?
10C8*2^8
Can you complete?
I will work it out thank you.

Another question. I have 30 green, 30 blue and 30 red balls. I must find the ways in which I can select 10 balls. I did this:

n=10
k=3 (basket of balls since all of them have 30 balls)

(n+k-1)! / k(n-1)!. Am I correct ?

Then I have 30 green, 30 blue and 5 red balls and must also find the ways to choose 10 . Now I must consider as k the number 30+30+5=65 and use the same type (n+k-1)!/k(n-1)! ?

thank you.

14. got it. thank you.

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