P&C q1

[JeffM]

Choosing 3 out of 5 balls--> 5c3

To use that in a computation, you need to simplify it. Where is that simplification?

Choosing 1 out of 2 balls --> 2c1

To use that in a computation, you need to simplify it. Where is that simplification?

Choosing 1 out of 1 --> 1c1

To use that in a computation, you need to simplify it. Where is that simplification?

Of course I know how to do them. My point is that you have not simplified them because you have not shown any computations with them so we have no clue what you are going to do with them.

Now for a particular combination (b1 b2 b3 , b4, b5)

Main point: For a event suppose, Triplets have three options of sitting in any one box so it chooses one , then there are two boxes left for 2 Single Distinct balls ; so ball 4 have 2 options so suppose b4 sits in any ONE out of 2 box and the remaining one box can be taken by b5 . So , 6 ways ?

Yes at last a computation with reasoning associated with it. The reasoning is not easy to follow but is quite sound. The computation is correct.

So, 5c3 * 2c1 * 1c1 * 3! . What is so hard to understand?

What is the mistake that am i not able to catch?[

You have not said what the indicated computation represents. That makes it impossible to understand. Moreover, because you have not told us what you are trying to calculate, how do we know if you made a mistake or not? Finally in no previous step have you given 3!. Where did that come from? And what does that expression simplify to?

You said your answer was 300. Now, according to my reckoning the expression above simplifies to

[MATH]\dbinom{5}{3} * \dbinom{2}{1} * \dbinom{1}{1} * 3! = 10 * 2 * 1 * 6 = 120 \ne 300.[/MATH]
So I have no idea how you got an answer of 300.

EDIT: By the way, 60 is a relevant number so the 120 does look as though you double counted something, but I do not know why you double counted because you did not tell us why you stuck those numbers together the way you did.

EDIT2: And I agree that combinatorics is hard. The first time I solved this problem I did it wrong. Only trying a different method showed me my error.

 

[Saumyojit]

But these two arrangements would also have resulted from choosing ACD, then B, then D. That is where you have double counted.

DID not understand.


5c3 --> B1 B2 B3 chosen out of 10 combo
2c1--> b4 chosen out of 2 combo (b4 or b5)
1c1---> B5

I have just chosen the combination of triplets and singles. I have not decide their positon yet.

Now deciding their positions , SO to arrange them among three boxes i am looking at these six arrangements below

(b1,b2,b3) , (b4), (b5)
(b1,b2,b3) , (b5), (b4)
(b4), (b5), (b1,b2,b3)
(b5), (b4), (b1,b2,b3)
(b4), ,(b1,b2,b3) , (b5)
(b5), (b1,b2,b3), (b4)


this is simple fpc right.


3options * 2options * 1 option
triplets single another single

(5c3 * 2c1 *1c1 ) --> this part is I have chosen what will be triplets and other two singles

When i am multiplying by Six , I am considering the six arrangements.

(5c3 * 2c1 *1c1 )* 6

 

[JeffM]

DID not understand.


5c3 --> B1 B2 B3 chosen out of 10 combo
2c1--> b4 chosen out of 2 combo (b4 or b5)
1c1---> B5

I have just chosen the combination of triplets and singles. I have not decide their positon yet.

Now deciding their positions , SO to arrange them among three boxes i am looking at these six arrangements below

(b1,b2,b3) , (b4), (b5)
(b1,b2,b3) , (b5), (b4)
(b4), (b5), (b1,b2,b3)
(b5), (b4), (b1,b2,b3)
(b4), ,(b1,b2,b3) , (b5)
(b5), (b1,b2,b3), (b4)


this is simple fpc right.

For any given triplet, there are six arrangements, we agree.

3c1 * 2c1 * 1c1 = 3 * 2 * 1 = 3 * 2 = 6. Simplify your expressions.

How many possible triplets are there? 5c3 = 10.

10 * 6 = 60. DONE.

You could list them in a table if you bothered to work the full table out rather than just one tenth of it.

(5c3 * 2c1 *1c1 ) --> this part is I have chosen what will be triplets and other two singles[

When i am multiplying by Six , I am considering the six arrangements.

(5c3 * 2c1 *1c1 )* 6

What you are missing, as I believe Dr. Peterson already explained to you several times, is this.

When you selected which box to put a GIVEN TRIPLETON in, you were left with a specific unique pair. There are only two distinct ways you can divide those two balls left between the two boxes left. That is where the two came from in calculating 6.

When you asked how many specific tripleton / pair combinations there were, your 5c3 = 10, there was no need for the 2c1 and 1c1. They were already accounted for in the 6 you calculated before. So you doubled the number by counting the pair possibilities when calculating the 6 and again when computing 5c3 * 2c1 * 1c1.

 

[Saumyojit]

Okay finally I think i understood .
THANKS for helping .

Concept that was used first : Division of Objects into equal and unequal groups

This is where I was mistaking.


5c3 --> B1 B2 B3 chosen out of 10 combo
2c1--> b4 chosen out of 2 combo (b4 or b5)
1c1---> B5

Here i was dividing 5 balls in 3 groups not boxes in 3:1:1
Now, group 2 and group 3 is the same thing only .

(B1B2B3) , B4 ,B5 and (B1B2B3) , B5 ,B4 --> These two are same . SO, 2c1 is overcounting .

so, final expression of Division into Groups = (5c3 * 2c1 * 1c1 )/ 2!

So, Now comes the distribution part -->

For a event suppose, Triplets have three options of sitting in any one box so it chooses one , then there are two boxes left for 2 Single Distinct balls ; so ball 4 have 2 options so suppose b4 sits in any ONE out of 2 box and the remaining one box can be taken by b5 . So , 6 ways .


3options * 2options * 1 option
triplets single another single


So, Final expression of (3,1,1) = (5c3 * 2c1 * 1c1 ) * 3

 

[JeffM]

Okay finally I think i understood .
THANKS for helping .

Concept that was used first : Division of Objects into equal and unequal groups

This is where I was mistaking.


5c3 --> B1 B2 B3 chosen out of 10 combo
2c1--> b4 chosen out of 2 combo (b4 or b5)
1c1---> B5

Here i was dividing 5 balls in 3 groups not boxes in 3:1:1
Now, group 2 and group 3 is the same thing only .

(B1B2B3) , B4 ,B5 and (B1B2B3) , B5 ,B4 --> These two are same . SO, 2c1 is overcounting .

so, final expression of Division into Groups = (5c3 * 2c1 * 1c1 )/ 2!

So, Now comes the distribution part -->

For a event suppose, Triplets have three options of sitting in any one box so it chooses one , then there are two boxes left for 2 Single Distinct balls ; so ball 4 have 2 options so suppose b4 sits in any ONE out of 2 box and the remaining one box can be taken by b5 . So , 6 ways .


3options * 2options * 1 option
triplets single another single


So, Final expression of (3,1,1) = (5c3 * 2c1 * 1c1 ) * 3

Yes, that solves half the problem correctly 10 * 2 * 1 * 3 = 60.

I think the way I explained it is more straightforward, but so long as your method always leads to correct answers, then any criticism is minor.

So give the other half a go.

 

[pka]

To Saumyojit Here in the use of inclusion/exclusion. If \(k=1,2,\text{ or }3\) let \(E_k\) be the event that box K is left empty.
We don't want to leave any box empty. So the number of ways at least one is empty is
\(\left| E_1\cup E_2\cup E_3 \right|=\left| E_1\cup E_2\cup E_3 \right|+\left| E_1\cap E_2 \right|+\left| E_2\cap E_3 \right|-\left| E_1\cap E_3 \right|+\left| E_1\cap E_2\cap E_3 \right|\) or
\(\left| E_1\cup E_2\cup E_3 \right|=2^5+2^5+2^5-1-1-1+0=93\)
But we want none empty so \(3^5-93=150\) Which is the number of onto functions from a set of five to a set of three.