DISTRIBUTING ITEMS/PEOPLE/NUMBERS... (QUESTION COLLECTION):

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hi

since chocolates are identical, and there are 3 children, we can think this way

c | c | c c c c

thus

8!
______
6! 2!

= 28, the answer
thanks

cheers, and do consider some kudos, guys

Hi

Multichoosing problem.

We need to distribute 6 identical chocolates among 3 children: (\(x_1, x_2, x_3\) - # each one will get)

\(x_1 + x_2 + x_3 = 6\)

\(_{3+6-1}C_6 = _8C_6 = _8C_2 = \frac{8*7}{2} = 28\)

Answer B

can be solve using stick method,,
imagine 2 sticks and 6 choclates ..
so.. it wud luk like this ||CCCCCC

permutation of the above = (8!/2! 6!) = 28

ans B

i took a bit long route -
3 +ve digit that can sum to 6
006-3 ways
051-6 ways
411-3 ways
321-6 ways
330- 3 ways
222- 1 way
204- 6 ways

total- 28 ways

Using stick method,

permutation of the above = (8!/2! 6!) = 28

So B

Let, three children get a, b and c chocolates respectively

i.e. a+b+c = 6

Now, the whole number solutions of an equation = (n+r-1)C(r-1) = (6+3-1)C(3-1) = 8C2 = 28

Answer: Option B

ALTERNATIVE

Let, three children get a, b and c chocolates respectively

i.e. a+b+c = 6

@a=0, (b,c) can be (0,6) (1,5) (2,4) (3,3) (4,2) (5,1) (6,0) i.e. 7 ways
@a=1, (b,c) can be (0,5) (1,4) (2,3) (3,2) (4,1) (5,0) i.e. 6 ways
@a=2, (b,c) can be (0,4) (1,3) (2,2) (3,1) (4,0) i.e. 5 ways
and so on...

Total Ways of distribution = 7+6+5+4+3+2+1 = 28

Answer: Option B


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Can someone explain the stick method, please?

Why 3+6-1? Can someone explain please?

Hello, can you please explain why the whole number solutions of the equation = (n+r-1)C(r-1)?
Thank you

Hi Guys,

If chocolates were different, I think that we could use the following Method

3*3*3*3*3*3 =3^6

Is there a way to move from this scenario to actual solution?

Could you please explain the rationale behind a different approach ?

Posted from my mobile device

This can not be corrected because this method includes some cases once, some cases TWICE and some cases thrice and the repetition in this case can not be eliminated

I hope this help!
Albymana
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Asked: In how many ways can 6 chocolates be distributed among 3 children? A child may get any number of chocolates from 0 to 6 and all the chocolates are identical.

C C C C C C ||

Number of ways = 8C2 = 28

IMO B
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Got your point thank you. The rationale behind is that

- when object are different you have to consider the single elements in the allocation (I.e. You consider two different ways when Child 1 receive C1, C2, C5 and child 2 receive C2, C1 and C5

- when object are identical, instead, you only have to care about # of element allocated to students

Am I right?

Best

It may seem odd at first, but this question can be reduced to the analogous question: In how many ways can we arrange the letters in IIOOOOOO?
Let me explain the relationship between this question and the original question.

First, however, let's say the three children are child A, child B, and child C

One possible arrangement of the 8 letters is OOIOIOOO
This arrangement represents child A receiving 2 chocolates, child B receiving 1 chocolate, and child C receiving 3 chocolates

Likewise, the arrangement OIOOOOIO represents child A receiving 1 chocolate, child B receiving 4 chocolates, and child C receiving 1 chocolate

And the arrangement OOIIOOOO represents child A receiving 2 chocolates, child B receiving 0 chocolates, and child C receiving 4 chocolates

Okay, enough examples.
Now that we understand the relationship between the original question and the analogous question, let's see how many ways we can arrange the letters in IIOOOOOO

----ASIDE--------------------
When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:
If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]
So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:
There are 11 letters in total
There are 4 identical I's
There are 4 identical S's
There are 2 identical P's
So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]
-------------------------------

In the case of the letters in IIOOOOOO, . . .
There are 8 letters in total
There are 2 identical I's
There are 6 identical O's
So, the total number of arrangements = 8!/(2!)(6!) = 28

Answer: B

Cheers,
Brent
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Just to get a better understanding:
This is not a permutation and instead it is a combination (permutation formula would be n!/(n-k)!), correct?

In this kind of problem 2 most common approaches:

1) straight formula (n + k - 1)! / (n - 1)! * k! ----> (3 + 6 - 1)! / (3 - 1)! * 6! = 8! / 6!*2! = 28

2) all possible groups:

6 - 0 - 0 3 ways
5 - 1 - 0 6 ways
4 - 2 - 0 6 ways
4 - 1 - 1 3 ways
3 - 3 - 0 3 ways
3 - 2 - 1 6 ways
2 - 2 - 2 1 way

TOTAL 28 ways

Trying to learn combinations and permutations. So the MISSISSIPPI rule says that if we had three different type of chocolates, two of each, the formula would have been

8!/(2!*2!*2!*2!) = 2520

is this correct?