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**niharika_kumar****Member**- From: Numeraland
- Registered: 2013-02-12
- Posts: 1,062

Q.) There are stall for 10 animals in a ship.In how many ways can the ship load be made if there are cows ,calves and horses to be transported, animals of each kind being not less than 10?

*Last edited by niharika_kumar (2014-03-31 19:59:25)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 100,247

Is there something missing in that space between transported and animals?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **Thinking is cheating.**

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**niharika_kumar****Member**- From: Numeraland
- Registered: 2013-02-12
- Posts: 1,062

sorry, that was a mistake.I corrected it.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Hi;

Is that the exact wording of the problem?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **Thinking is cheating.**

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**niharika_kumar****Member**- From: Numeraland
- Registered: 2013-02-12
- Posts: 1,062

yes and i find it difficult to understand.

the answer in the book is given as 3^10.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 100,247

There wording is a little funny but there are 3 for the first stall and 3 for the second and 3 for the third... 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 = 3^10.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **Thinking is cheating.**

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**niharika_kumar****Member**- From: Numeraland
- Registered: 2013-02-12
- Posts: 1,062

i still don't get this.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Here is one permutation of the stalls:

{cow, cow, calf, horse, cow, horse, horse, calf, calf, cow}

here is another

{horse, cow, calf, cow, cow, cow, cow, horse, calf, cow}

and another

{cow, cow, horse, calf, horse, calf, calf, cow, cow, calf}.

Do you see that each stall has 3 choices, a cow or a horse or a calf?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **Thinking is cheating.**

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**niharika_kumar****Member**- From: Numeraland
- Registered: 2013-02-12
- Posts: 1,062

so it was all about this.

"animals of each kind being not less than 10" <---what was the use of this sentence then

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Since the title is 'permutation', I am assuming that the order matters.

For Each stall , you can do it in 3 ways (i.e, one of cows or calves or horses)

For 10 stalls, you can do it in 3^10 Ways

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

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niharika_kumar wrote:

so it was all about this.

"animals of each kind being not less than 10" <---what was the use of this sentence then

I do not know but who cares?

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**niharika_kumar****Member**- From: Numeraland
- Registered: 2013-02-12
- Posts: 1,062

hmmm, this sentence confused me.

thanks.

friendship is tan 90°.

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It means that the pool from which you are choosing animals has atleast 10 of each of those animals

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 100,247

hmmm, this sentence confused me.

Confused me too.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **Thinking is cheating.**

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**eigenguy****Member**- Registered: 2014-03-18
- Posts: 78

I believe the intent is:

The ship can be loaded with 10 animals.

Animals available to be loaded consist of three types: Cows, calves, horses.

There are enough of each type of animal available to fill all 10 stalls with cows, or all 10 with calves, or all 10 with horses, if you desired.

Individual cows are not distinguishable from each other (so exchanging one cow for another does not constitute a different loading). The same is true of calves, and of horses.

However, and I base this only on the answer provided, not something that made it clear in the problem, the individual stalls are distinguished (so having a cow in #1 and a horse in #2 is a different loading from having a horse in #1 and a cow in #2, even if all the other stalls are the same).

Based on that understanding, the type of animal in each stall is independent from the types in the other stalls, and there are exactly 3 ways to fill each stall, so the total number of loadings is 3*3*3*...*3 = 3^10.

"Having thus refreshed ourselves in the oasis of a proof, we now turn again into the desert of definitions." - Bröcker & Jänich

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**niharika_kumar****Member**- From: Numeraland
- Registered: 2013-02-12
- Posts: 1,062

it makes it much clearer.

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