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# if there are 10 elephants and 10 homes for elephants, how many elephants will go in each home?

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if there are 10 elephants and 10 homes for elephants, how many elephants will go in each home?

Jun 5, 2015

#3
+4682
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Melody is correct,

It depends on how many elephants the homes can hold.

It can be 2 , 3 , 1 , 7 ...?

Jun 6, 2015

#1
+600
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1 elephant in each home.

Jun 6, 2015
#2
+95177
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idk Eloise, Maybe one home gets all 10 elephants  ://     LOL

Jun 6, 2015
#3
+4682
+11

Melody is correct,

It depends on how many elephants the homes can hold.

It can be 2 , 3 , 1 , 7 ...?

MathsGod1 Jun 6, 2015
#4
+600
+8

Hmm, I didn't think about it that way before. You're right MathsGod. You too Miss Melody.

Jun 6, 2015
#5
+95177
+8

I could make a counting question out of this!

There are 10 elephants that I will class as identical, and 10 different homes that they can go to.

Any number of elphants can go to any of the homes.  How many ways can this be done?

The bars represent the homes and the * represent the elephants.

The elephants are entering the door of the hous from the left side.

This would  be one combination

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

1 elephant in house 1

3 elephants in house 2

NO elephants in house 3 or 4

2 elephants in house 5

No elephant in house 6

2 elephants in house 7

2 elephants in house 8

No elephants in house 9 or 10

Now if I put the stars in different spots the the number of elephants in one or mor houses will be different!

The last bar has to be at the end so its position cannot change and it is not used in the calculation.

Altogether there are 19 stars and bars left and I want to know how many ways I can choose 10 of those to be elephants (stars)

That is 19C10

$${\left({\frac{{\mathtt{19}}{!}}{{\mathtt{10}}{!}{\mathtt{\,\times\,}}({\mathtt{19}}{\mathtt{\,-\,}}{\mathtt{10}}){!}}}\right)} = {\mathtt{92\,378}}$$

I think that is right.  There are a lot of different ways our elephants can be housed!!

NOTE:  I could look at this a little differently

Altogether there are 19 stars and bars left and I want to know how many ways I can choose 9 of those to be houses (bars)

This would be    19C9

$${\left({\frac{{\mathtt{19}}{!}}{{\mathtt{9}}{!}{\mathtt{\,\times\,}}({\mathtt{19}}{\mathtt{\,-\,}}{\mathtt{9}}){!}}}\right)} = {\mathtt{92\,378}}$$

See, the answer is the same!!

Jun 6, 2015
#6
+4682
0

.
Jun 6, 2015
#7
+600
0

Woah!

I would never be able to do all that! It makes my head spin! I don't get how mathematic teachers come up with all these solutions!

Jun 6, 2015
#8
+95177
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If you study maths for long enough you will learn this stuff too Eloise :))

Jun 6, 2015
#9
+600
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I fear I will have to Miss Melody, for I aim to be a vet one day!  Lol.

Jun 6, 2015
#10
+95177
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Well Eloise you had better know how many ways you can house your elephants then!!      LOL

Jun 6, 2015
#11
+4682
0

LOL!

It's true:

Jun 6, 2015