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## Comment on

Arranging Buttons## Here can we approach this way

Since total no. of buttons are 6

Red buttons =2

Blue buttons =2

Green buttons = 2

Therefore the no. of ways = 6C2 * 4C2 * 2C2 =90 ways

## It took me a while to

It took me a while to determine what you are doing with each step in your solution.

I think I understand now.

There are 6 places each button can go.

So, first we can choose two spaces to place the two red buttons.

We can accomplish this in 6C2 ways.

There are now 4 spaces left.

Next, we'll choose two spaces to place the blue buttons.

We can accomplish this in 4C2 ways.

There are now 2 spaces left.

Finally, we'll choose two spaces to place the green buttons.

We can accomplish this in 2C2 ways.

Great strategy!!!

Cheers,

Brent

## Another great approach.

## Apology, just dashed

## I'm confused because I

## Those two questions, although

Those two questions, although similar, have a big difference.

In the books on a shelf question, the books in one topic (e.g, History) are DIFFERENT. For example, one history book might be about the Civil War, which another history book might be about the War of 1812

In the above question, the two red buttons are IDENTICAL, the two blue buttons are IDENTICAL, and the two green buttons are IDENTICAL.

Cheers,

Brent

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