# Permutations with different numbers of objects

• Mar 24th 2013, 07:14 AM
qwertyasd
Permutations with different numbers of objects
Hi there.

I am trying to find the number of different permutations for the following problem:

 Box Box 1 Red Blue Green Box 2 Red Blue Green Box 3 Red Blue Green Box 4 Red Blue Green Box 5 Red Blue n/a Box 6 Red Blue n/a Box 7 Red Blue n/a

I have seven boxes.Three of the boxes (box 5,6&7) can be either red or blue. Four of the boxes (box 1,2,3&4) can be either red, blue or green. How do I find out the number of permutations of all seven boxes?

i.e
What is the value of n?
 Combination Box 1 Box 2 Box 3 Box 4 Box 5 Box 6 Box 7 Combination 1 R R R R R R R Combination 2 R R R R R R B Combination 3 R R R R R B R ..... ... ... ... ... ... ... ... Combination n

Thanks
• Mar 24th 2013, 08:00 AM
majamin
Re: Permutations with different numbers of objects
Can you answer the following (easier, but analogous) question? I have 3 different pairs of shirts and 2 different pairs of pants. How many different outfits can I make? How can this help you answer your question?
• Mar 24th 2013, 08:29 AM
qwertyasd
Re: Permutations with different numbers of objects
hi,

Thanks for responding.
"I have 3 different pairs of shirts and 2 different pairs of pants. How many different outfits can I make?"

I believe that you could make 6 different outfits.

Im not sure how this helps me?
• Mar 24th 2013, 01:31 PM
majamin
Re: Permutations with different numbers of objects
You have three choices for Box 1, 2, 3, and 4 and two choices for 5, 6, and 7. All are independent choices, therefore, it is similar to choosing outfits, just as you calculated (i.e. the product of choices).
• Mar 25th 2013, 12:59 AM
qwertyasd
Re: Permutations with different numbers of objects
I understand now.

So
n= (3^4)*(2^3)
= 648

If so, thanks!
• Mar 25th 2013, 09:54 AM
majamin
Re: Permutations with different numbers of objects
Quote:

Originally Posted by qwertyasd
I understand now.

So
n= (3^4)*(2^3)
= 648

If so, thanks!

That's correct.