# 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 Note by Pratik Patel
10 months, 2 weeks ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
• Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears 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} $\boxed{123}$

## Comments

Sort by:

Top Newest

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

- 10 months, 2 weeks ago

Log in to reply

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

- 10 months, 2 weeks ago

Log in to reply

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

This is the question

- 10 months, 2 weeks ago

Log in to reply

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.

- 10 months, 2 weeks ago

Log in to reply

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

- 10 months, 2 weeks ago

Log in to reply

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.

- 10 months, 2 weeks ago

Log in to reply

What will be the answer?

- 10 months, 2 weeks ago

Log in to reply

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 $3$.

- 10 months, 2 weeks ago

Log in to reply

What will be the answer?

- 10 months, 2 weeks ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...