# Inclusion Exclusion

• Mar 8th 2011, 07:58 PM
stumped765
Inclusion Exclusion
How many ways to arrange the letters of SCIENCE BRAIN
such that there are exactly 3 sets of the same letter conseculatively.

the total number of ways when doing regularily is 14!/2^4

since the question says that it wants 3 sets of the same letter conseculatively im thinking i would find that there are 13!/2^3 ways to have 1 set of the same letter consecutatively...except that just blocks one set of letters and doesnt garuntee for the other letters..and normally when doing PIE the cases are the opposite of the question so should i be finding the ways such that they are not consecutive?
• Mar 9th 2011, 02:49 AM
Plato
Quote:

Originally Posted by stumped765
How many ways to arrange the letters of SCIENCE BRAIN
such that there are exactly 3 sets of the same letter conseculatively.
the total number of ways when doing regularily is 14!/2^4

There are four ways to select three pairs to be together.
Then there are \$\displaystyle \dfrac{11!}{2}\$ ways to have those three pairs together.
But that also counts the case in which all four pairs are together. Thus subtract that case.
• Mar 9th 2011, 10:21 AM
stumped765
Quote:

Originally Posted by Plato
There are four ways to select three pairs to be together.
Then there are \$\displaystyle \dfrac{11!}{2}\$ ways to have those three pairs together.
But that also counts the case in which all four pairs are together. Thus subtract that case.

if we had sCCIINNeebr thats 3 pairs, so the number of ways to have that.. wouldnt that be 8!/2? since the CC II NN act as blocks?

This is what i have
4C3*8!/2 - 4C4*7!
• Mar 9th 2011, 10:38 AM
Plato
Quote:

Originally Posted by stumped765
if we had sCCIINNeebr thats 3 pairs, so the number of ways to have that.. wouldnt that be 8!/2? since the CC II NN act as blocks? This is what i have 4C3*8!/2 - 4C4*7!

I took your own count. But there are 12 letters not 14.