Sample example in combination

Hi,

There is sample example in combination which states that how many ways 9 chairs can be order in the group of 3 chairs?

Some part I have understood but in the answer it is divided by 3! My question is why divided by 3!.

REGARDS

#Logic

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
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Comments

Could it be because they asked for the number of combinations instead of permutations? Go here for an explanation of combinations.

David Stiff - 10 months, 1 week ago

Or there could be repeats. I'd have to see the problem to know for sure.

David Stiff - 10 months, 1 week ago

There are 99 distinct chairs. How many ways are there to group these chairs into 3 groups of 3?

This is the question

Pratik Patel - 10 months, 1 week ago

Oh, I see now. Did you mean $9$ chairs? If so, then the first step, which it sounds like you understand, would be to find $9!$ ( $9$ chairs to pick, then $8$ chairs, and so on). However, we need to divide this by the number of ways we can arrange the groups of $3$ themselves.

For instance, one way to arrange the chairs could be $ABC$ , $DEF$ , and $GHI$ . However, if we don't divide by the number of group arrangements, we will also be counting other arrangements like $GHI$ , $DEF$ , and $ABC$ , which is the same as our first one, just with the groups rearranged. We don't consider these to arrangements to be distinct, since each group is the same, just ordered differently. Thus, we have to divide by the number of ways we could arrange the groups, in our case, $3!$ .

Hope this helps.

David Stiff - 10 months, 1 week ago

Thanks for the reply..

Suppose We have 6 people and we have to make 2 groups and each group contains 3 people. So in this case do we have to divide by 2! Or 3! And what will be the answer?

Regards

Pratik Patel - 10 months, 1 week ago

Good question! In this case we would have to divide by $2!$ , since there are $2$ groups, and there would be $2!$ ways to arrange them.

It's kind of like each time we come up with another way to distribute the people, we also have $2!$ ways to order the groups themselves. So we have to divide by $2!$ to only consider the distribution itself.

David Stiff - 10 months, 1 week ago

What will be the answer?

Pratik Patel - 10 months, 1 week ago

@Pratik Patel – Oh, sorry. I didn't see that part of your question. The answer would be $\frac{6!}{2!} = 6\cdot5\cdot4\cdot3 = \boxed{360}$ ways to arrange $6$ people in $2$ groups of $3$ .

David Stiff - 10 months, 1 week ago

What will be the answer?

Pratik Patel - 10 months, 1 week ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$