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Order does not matter--multiply or divide

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Order does not matter--multiply or divide

by 2mist » Sat Jan 18, 2014 8:56 pm
Hi Brent,

I am not sure how to put my doubt in words, but I hope you will get a faint idea of what I am talking about and help me, and alike, out in understanding the concept.

1. Some times when we say order doesn't matter, we divide the the factorial : -

eg:

picking 2 members from a group of 5 (A, B, C, D, E) members :

(5x4)/ 2 we divide by 2 here because A and B can occur in any order i.e AB and BA both are same.


2. Some times when we have to take into account that items can appear in any order, we multiply a factorial ( doing so, according to me, also means order doesn't matter)

eg.

if we have 5 different balls to distribute in 3 different baskets we can do it in following ways

(3 * 1 * 1) *3---here we are multiplying the factorial by 3 because order doesn't matter and 3 balls can be placed in any of the basket.

or

(2*2*1)*3 ---here we are multiplying the factorial by 3 because order doesn't matter and 2 balls can be placed in any of the 2 baskets out of 3.

There is some huge confusion here. Please help me understand the logic.

Thanks,
Mist
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by brianlange77 » Sat Jan 18, 2014 9:14 pm
2mist wrote:Hi Brent,

I am not sure how to put my doubt in words, but I hope you will get a faint idea of what I am talking about and help me, and alike, out in understanding the concept.

1. Some times when we say order doesn't matter, we divide the the factorial : -

eg:

picking 2 members from a group of 5 (A, B, C, D, E) members :

(5x4)/ 2 we divide by 2 here because A and B can occur in any order i.e AB and BA both are same.


2. Some times when we have to take into account that items can appear in any order, we multiply a factorial ( doing so, according to me, also means order doesn't matter)

eg.

if we have 5 different balls to distribute in 3 different baskets we can do it in following ways

(3 * 1 * 1) *3---here we are multiplying the factorial by 3 because order doesn't matter and 3 balls can be placed in any of the basket.

or

(2*2*1)*3 ---here we are multiplying the factorial by 3 because order doesn't matter and 2 balls can be placed in any of the 2 baskets out of 3.

There is some huge confusion here. Please help me understand the logic.

Thanks,
Mist
Mist -- to address your question, let me ask you this. Is your question about the mathematical steps themselves? Or is it about understanding the situations when 'order' does matter vis a vis those situations when it doesn't matter? More than happy to help either way -- but wanted to make sure I (or others) can answer your REAL question!
Thanks,
Brian
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by 2mist » Sat Jan 18, 2014 9:34 pm
brianlange77 wrote:
2mist wrote:Hi Brent,

I am not sure how to put my doubt in words, but I hope you will get a faint idea of what I am talking about and help me, and alike, out in understanding the concept.

1. Some times when we say order doesn't matter, we divide the the factorial : -

eg:

picking 2 members from a group of 5 (A, B, C, D, E) members :

(5x4)/ 2 we divide by 2 here because A and B can occur in any order i.e AB and BA both are same.


2. Some times when we have to take into account that items can appear in any order, we multiply a factorial ( doing so, according to me, also means order doesn't matter)

eg.

if we have 5 different balls to distribute in 3 different baskets we can do it in following ways

(3 * 1 * 1) *3---here we are multiplying the factorial by 3 because order doesn't matter and 3 balls can be placed in any of the basket.

or

(2*2*1)*3 ---here we are multiplying the factorial by 3 because order doesn't matter and 2 balls can be placed in any of the 2 baskets out of 3.

There is some huge confusion here. Please help me understand the logic.

Thanks,
Mist
Mist -- to address your question, let me ask you this. Is your question about the mathematical steps themselves? Or is it about understanding the situations when 'order' does matter vis a vis those situations when it doesn't matter? More than happy to help either way -- but wanted to make sure I (or others) can answer your REAL question!
Thanks,
Brian

Hi Brian,

Thanks for your quick response.

I don't have problem understanding the mathematical steps involved in solving either of the question. I understand when to use combination and when to use permutation, Yet I feel there is some gap.

I don't understand why sometimes we multiply to take into account that the ball, or for that matter any thing, can appear in either of the three places.
eg.
3 balls can be placed in 1st or 2nd or 3rd basket so we multiply by 3.

where as, we generally apply combination to take into account that order doesn't matter, dividing the factorial rather than multiplying.

May be a few questions, with good explanation, where we use the latter will hep solidify my understanding, coz i am confident about combination technique.

Regards,
Mist
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by 2mist » Sat Jan 18, 2014 9:46 pm
2mist wrote:
brianlange77 wrote:
2mist wrote:Hi Brent,

I am not sure how to put my doubt in words, but I hope you will get a faint idea of what I am talking about and help me, and alike, out in understanding the concept.

1. Some times when we say order doesn't matter, we divide the the factorial : -

eg:

picking 2 members from a group of 5 (A, B, C, D, E) members :

(5x4)/ 2 we divide by 2 here because A and B can occur in any order i.e AB and BA both are same.


2. Some times when we have to take into account that items can appear in any order, we multiply a factorial ( doing so, according to me, also means order doesn't matter)

eg.

if we have 5 different balls to distribute in 3 different baskets we can do it in following ways

(3 * 1 * 1) *3---here we are multiplying the factorial by 3 because order doesn't matter and 3 balls can be placed in any of the basket.

or

(2*2*1)*3 ---here we are multiplying the factorial by 3 because order doesn't matter and 2 balls can be placed in any of the 2 baskets out of 3.

There is some huge confusion here. Please help me understand the logic.

Thanks,
Mist
Mist -- to address your question, let me ask you this. Is your question about the mathematical steps themselves? Or is it about understanding the situations when 'order' does matter vis a vis those situations when it doesn't matter? More than happy to help either way -- but wanted to make sure I (or others) can answer your REAL question!
Thanks,
Brian

Hi Brian,

Thanks for your quick response.

I don't have problem understanding the mathematical steps involved in solving either of the question. I understand when to use combination and when to use permutation, Yet I feel there is some gap.

I don't understand why sometimes we multiply to take into account that the ball, or for that matter any thing, can appear in either of the three places.
eg.
3 balls can be placed in 1st or 2nd or 3rd basket so we multiply by 3.

where as, we generally apply combination to take into account that order doesn't matter, dividing the factorial rather than multiplying.

May be a few questions, with good explanation, where we use the latter will hep solidify my understanding, coz i am confident about combination technique.

Regards,
Mist

Hi Brian,

Lets shelve combinations for a while, I feel i know when to use it.

Please help me understand the logic behind multiplying the factorial with with 3! to take into account that things can appear in 1st place or 2nd place or 3rd place any of the places.

Also, A few link that discuss such questions will be very much helpful.

Regards,
Mist
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by Brent@GMATPrepNow » Sat Jan 18, 2014 9:51 pm
Hi Mist,

I'd like to jump in and mention that I recently wrote an article on this topic: https://www.beatthegmat.com/mba/2013/09/ ... s-part-iii
This article is part III in a 3-part series examining counting techniques.

Here are parts I and II as well:
- https://www.beatthegmat.com/mba/2013/07/ ... ons-part-i
- https://www.beatthegmat.com/mba/2013/08/ ... ns-part-ii

Cheers,
Brent
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by 2mist » Sun Jan 19, 2014 5:13 am
Brent@GMATPrepNow wrote:Hi Mist,

I'd like to jump in and mention that I recently wrote an article on this topic: https://www.beatthegmat.com/mba/2013/09/ ... s-part-iii
This article is part III in a 3-part series examining counting techniques.

Here are parts I and II as well:
- https://www.beatthegmat.com/mba/2013/07/ ... ons-part-i
- https://www.beatthegmat.com/mba/2013/08/ ... ns-part-ii

Cheers,
Brent
HI Brent,

I have been using FCP concepts for solving counting problem and they have indeed made my life easier. However, your articles on this topic are awesome and have solidified my understanding of this topic even further.

"Does the outcome of one stage differ from the outcome of the other stage?" -- did give me an aha moment today..i will make sure I narrate my kids how uncle Brent helped their daddy get into a MBA college, helping meet their mother and thus in a way, are responsible for their existence. :P


Having said that, I do have few doubts and will greatly appreciate your input.

1.
From the set of integers from 1 to 100 inclusive, in how many ways can we select two different numbers such that their sum is an EVEN integer?

To solve this shouldn't we consider two cases:

a. select an even number * select an odd number = 50*50 = 2500
OR
b. select an odd number * select an even number = 50*50 = 2500

answer = 5000 ( you have derived the answer 2500, considering only case 'a' )

2. If my above understanding is wrong, then what are the situations in which we apply such cases?


Please help me solve these two problems in your approach. I don't have problem in understanding the steps involved; however, I believe by reading your solution I will gain further insight.

3. In how many ways you can distribute 5 marbles in 3 identical baskets such that each basket has at least one marble.

4. In how many ways you can distribute 5 different coloured marbles in 3 distinct baskets such that each basket has at least one marble.

Regards,
Mist
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by Brent@GMATPrepNow » Sun Jan 19, 2014 8:57 am
2mist wrote: i will make sure I narrate my kids how uncle Brent helped their daddy get into a MBA college, helping meet their mother and thus in a way, are responsible for their existence. :P
I hope I'm invited to the wedding!
Also, "Brent" is a lovely name for both boys and girls. :-)

2mist wrote:
1.
From the set of integers from 1 to 100 inclusive, in how many ways can we select two different numbers such that their sum is an EVEN integer?

To solve this shouldn't we consider two cases:

a. select an even number * select an odd number = 50*50 = 2500
OR
b. select an odd number * select an even number = 50*50 = 2500

answer = 5000 ( you have derived the answer 2500, considering only case 'a' )
Your solution has two issues.
First, if the integers are different, then you only have 49 choices for the second number.
Also, "50*50" suggests that you're treating the outcomes as different, when they are the same.
This solution requires combinations.

Here are two valid solutions.

Solution #1:
# of ways to select 2 EVEN numbers from 50 EVEN numbers = 50C2 = 1225
# of ways to select 2 ODD numbers from 50 ODD numbers = 50C2 = 1225
Total = 1225 + 1225 = 2450

Solution #2 (tricky):
Let's first IGNORE the condition that the two selected numbers must have an even sum.
We can select 2 numbers from 100 numbers in 100C2 ways (= 4950 ways)
Of course some of those selections will have an odd sum and some will have an even sum.
IMPORTANT: Half of the 4950 selections will have an even sum.
So, the number of selections with an even sum = 4950/2 = 2450
2mist wrote:
3. In how many ways you can distribute 5 marbles in 3 identical baskets such that each basket has at least one marble.

4. In how many ways you can distribute 5 different coloured marbles in 3 distinct baskets such that each basket has at least one marble.

Regards,
Mist
I need to leave the forums for a while, so it might be useful if you post these as two separate questions.

Cheers,
Brent
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by 2mist » Tue Jan 21, 2014 11:20 am
Hello Brent,

I believe that i am missing something in the solution that you have provided above.

You have solved the same thing here - https://www.beatthegmat.com/mba/2013/09/ ... s-part-iii

Now let's try a harder question.

From the set of integers from 1 to 100 inclusive, in how many ways can we select two different numbers such that their sum is an odd integer?

Let's first see what happens when we ask, "Does the order of the two selected numbers matter?"

Well, selecting 7 then 20 (for an odd sum of 27) is the same as selecting 20 then 7. Since the order doesn't matter, this must be a combination question, right?

No. It turns out that this is not a combination question, even though it appears that order doesn't matter.

Now let's see what happens when we apply the FCP.

We'll take the task of selecting 2 numbers (with an odd sum), and we'll break it into stages. One way to do this is to first recognize that, for the sum to be odd, one number must be even, and the other number must be odd. So, our two stages could look like this:

Stage 1: Select an even number from the set

Stage 2: Select an odd number from the set

Does the outcome of one stage differ from the outcome of the other stage?

YES. In one stage we're selecting an even number, and in the other stage, we're selecting an odd number. Since the outcomes are different, we can continue applying the FCP.

We can complete stage 1 in 50 ways (since there are 50 even numbers in the set), and we can complete stage 2 in 50 ways (since there are 50 odd numbers in the set). By the FCP, we can complete both stages (and thus select 2 numbers with an odd sum) in (50)(50) ways ( = 2500 ways).
I am pretty much convinced with this solution; however i fail to understand why didn't we consider two cases i.e
a. choosing a odd number then an even number.
b. choosing an even number then an odd number.

doesn't the solution 2500 is answer for question how many ways can we select two different numbers, first even then odd, such that their sum is an odd integer.


Also, as you suggested, I have posted my other two problems in below mentioned link.

https://www.beatthegmat.com/coloured-mar ... tml#708413
https://www.beatthegmat.com/marbles-comb ... tml#708412

#Thank you for giving awesome insights :)

Regards
Mist
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by Brent@GMATPrepNow » Tue Jan 21, 2014 5:00 pm
2mist wrote:Hello Brent,

I believe that i am missing something in the solution that you have provided above.

You have solved the same thing here - https://www.beatthegmat.com/mba/2013/09/ ... s-part-iii

Now let's try a harder question.

From the set of integers from 1 to 100 inclusive, in how many ways can we select two different numbers such that their sum is an odd integer?

Let's first see what happens when we ask, "Does the order of the two selected numbers matter?"

Well, selecting 7 then 20 (for an odd sum of 27) is the same as selecting 20 then 7. Since the order doesn't matter, this must be a combination question, right?

No. It turns out that this is not a combination question, even though it appears that order doesn't matter.

Now let's see what happens when we apply the FCP.

We'll take the task of selecting 2 numbers (with an odd sum), and we'll break it into stages. One way to do this is to first recognize that, for the sum to be odd, one number must be even, and the other number must be odd. So, our two stages could look like this:

Stage 1: Select an even number from the set

Stage 2: Select an odd number from the set

Does the outcome of one stage differ from the outcome of the other stage?

YES. In one stage we're selecting an even number, and in the other stage, we're selecting an odd number. Since the outcomes are different, we can continue applying the FCP.

We can complete stage 1 in 50 ways (since there are 50 even numbers in the set), and we can complete stage 2 in 50 ways (since there are 50 odd numbers in the set). By the FCP, we can complete both stages (and thus select 2 numbers with an odd sum) in (50)(50) ways ( = 2500 ways).
I am pretty much convinced with this solution; however i fail to understand why didn't we consider two cases i.e
a. choosing a odd number then an even number.
b. choosing an even number then an odd number.

doesn't the solution 2500 is answer for question how many ways can we select two different numbers, first even then odd, such that their sum is an odd integer.


Also, as you suggested, I have posted my other two problems in below mentioned link.

https://www.beatthegmat.com/coloured-mar ... tml#708413
https://www.beatthegmat.com/marbles-comb ... tml#708412

#Thank you for giving awesome insights :)

Regards
Mist
Hi Mist,

Sorry for the delay - I was away for a couple of days.
There's a small problem with your solution (in bold blue)

The two cases you suggested (i.e., selecting odd number then an even number AND selecting an even number then an odd number) are not different.
For example, selecting a 2 then a 7, is the same as selecting a 7 then a 2.


Cheers,
Brent
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