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[obliviousninja]

2012 Q16a)

ii) In how many ways can 10 identical coins be allocated to 4 different boxes? [looks easier than it actually is - Q16]

 

[Obvious]

Imagine lining up the coins as so:

* * * * * * * * * *

Allocating them to different boxes amounts to placing three dividers between the boxes, where the group of coins between each divider and on the sides are placed in unique boxes.

* * * | * * | * * * * | *

(3, 2, 4, 1)

We have 9 choices for the dividers and three dividers, so the answer should be 9C3 (assuming we can't have empty boxes).

If we can have empty boxes, then the answer's slightly different (I'll edit in a sec).

Edit:

This time let the dots just be placeholders. We line up thirteen:

* * * * * * * * * * * * *

And then select three to be the dividers (13C3). The dots between the dividers then become coins and are placed in unique boxes.

| | * * * * * * * * * | *

(0, 0, 9, 1)

 

[obliviousninja]

The answer is 286

 

[Sy123]

the answer is 286

13c3 = 286

 

[obliviousninja]

13c3 = 286

explanation please.

 

[Menomaths]

explanation please.

Calculat0r

 

[obliviousninja]

Calculat0r

nss, how did u get 13c3

 

[Sy123]

explanation please.

Imagine lining up the coins as so:

* * * * * * * * * *

Allocating them to different boxes amounts to placing three dividers between the boxes, where the group of coins between each divider and on the sides are placed in unique boxes.

* * * | * * | * * * * | *

(3, 2, 4, 1)

We have 9 choices for the dividers and three dividers, so the answer should be 9C3 (assuming we can't have empty boxes).

If we can have empty boxes, then the answer's slightly different (I'll edit in a sec).

Edit:

This time let the dots just be placeholders. We line up thirteen:

* * * * * * * * * * * * *

And then select three to be the dividers (13C3). The dots between the dividers then become coins and are placed in unique boxes.

| | * * * * * * * * * | *

(0, 0, 9, 1)

He edited his answer

 

[Drongoski]

as explained by Obvious

or = number of ways of arranging 13 objects, with 10 of 1 type (coins) and 3 of another (stick dividers).

like the number of ways of arranging the 10 letters in PARRAMATTA (2xR, 2xT, 4xA) = 10!/[2!2!4!]