Question

In how many ways can seven identical candies distributed to 3 children {I, II, III} if each child can get any number of candies?

Answer 1

isnt that just a count in base 7?

000
001
002
...
007
010
011
012
....

or do all the candies have to be handed out at once?

Answer 2

i spose it would have to be a little modified for to exclude stuff like 117 since that adds to 9

Answer 3

What is so special when that adds to 9?

Answer 4

I suppose all the candies will be handed out...

Answer 5

n-1 C r-1 r is kids and n is candies

Answer 6

@divu.mkr Can you explain it?

Answer 7

this is same as, finding # of ordered pairs of whole numbers x, y, z such that

x+y+z = 7

Answer 8

u familiar wid stars and bars ?

Answer 9

Not quite.

Answer 10

"Suppose one has k objects (to be represented as stars; in the example below k = 7) to be placed into n bins (in the example n = 3), such that all bins contain at least one object;"

But in this case, it is possible that the bin contains no objects.

Answer 11

just remember this formula when dividing n similar things to r persons/variables

Answer 12

But why it works in this case?

Answer 13

because identical and similar means same!! :D

Answer 14

But why n-1 and r-1 ?

Answer 15

that what the formula is.... and try to memorize it rather thn derivation

Answer 16

Maths is not just about remember the formula. Even though I remember it, I don't know when I can apply it, that means just memorizing is useless.

Btw, the formula is for all children having at least one candy. However, it is possible that some children do not even have a candy. How can it be applied here?

Answer 17

(Add the special case to the formula lol)

Answer 18

if u knew that then, y did you putted up ur ques here..?

Answer 19

I didn't know....

Answer 20

But apart from using that stars and bars formula with adding the special case, how to solve this question. (I am not even sure if using stars and bars formula with adding the special case works.)

Answer 21

stars and bars works perfectly here,

7 stars, 2 bars :-
```
**|***|**
```

Answer 22

the \(2\) bars can be placed in available \(7+2 = 9\) slots in \(\large \binom {9}{2} = 36\) ways

Answer 23

So # of whole number solutions for the equation, x+y+z = 7 wud be 36

Answer 24

What about ||******* ? Has it been counted?

Answer 25

yup bars can be anywhere

Answer 26

```
||*******

*******||
```

Answer 27

think of 7 stars + 2 bars as slots to be filled.

so 9 C 2 accounts for, ways to choose 9 slots for 2 bars

Answer 28

to convince ourselves, we can try below simple problem :-

x+y = 7

Answer 29

goal is to find all whole number x, y such that , they addup to 7

Answer 30

we set it up using, 7 stars and 1 bar

```
****|**
```

8 slots, and 1 bar

so # of possible solutions is 8 C 1 = 8