Posted by **navroz** on Tuesday, January 3, 2012 at 2:12pm.

the cookie monster has a package of cookies with him consisting of 4 chocolate chip, 5 raisin and 6 almond nut, 7 other different assorted cookies. if he reaches into the package and eats all the cookies, eating one cookie at a time, how many different eating orders are there?

- math -
**Steve**, Tuesday, January 3, 2012 at 2:31pm
4+5+6+7 = 22

So, there would be

22!

-------

4!5!6!

= 542,052,820,108,800

- math -
**navroz**, Tuesday, January 3, 2012 at 2:43pm
but why is the denominator 4!5!6!

- math -
**Steve**, Tuesday, January 3, 2012 at 2:53pm
If all the cookies were different, you'd have just 22! different ways. But, if you just consider the 4 chocolate chip cookies, they are indistinguishable from each other. If they were different, then there would be 4! different ways to eat the chocolate chip cookies. But you don't care which of them you are eating, so you divide by the 4! ways which are the same.

Consider the case of 4 cookies, 3 of which are A, and also a single B. If you write down all the ways to eat them, numbering the A's, you get

A1 A2 A3 B

A1 A2 B A3

A1 A3 A2 B

A1 A3 B A2

A1 B A2 A3

A1 B A3 A2

A2 A1 A3 B

A2 A1 B A3

A2 A3 A1 B

A2 A3 B A1

A2 B A1 A3

A2 B A3 A1

A3 A1 A2 B

A3 A1 B A2

A3 A2 A1 B

A3 A2 B A1

A3 B A1 A2

A3 B A2 A1

B A1 A2 A3

B A1 A3 A2

B A2 A1 A3

B A2 A3 A1

B A3 A1 A2

B A3 A2 A1

As you can see, there are 4! = 24 ways to eat the cookies. But, if all the A1 A2 A3 are replaced by just A, we have only

A A A B

A A B A

A B A A

B A A A

which is 4!/3! = 4

Expand that to your problem, and you see why we divide by n! if there are n identical objects.

- math -
**navroz**, Tuesday, January 3, 2012 at 3:11pm
but wat happened to the 7 shouldnt it be 4!5!6!7!

- math -
**Steve**, Tuesday, January 3, 2012 at 3:12pm
There are 7 assorted other cookies. Thus, they are different, and can be eaten in distinguishable orders.

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