# Thread: Counting using inclusion/exclusion principle.

1. ## Counting using inclusion/exclusion principle.

I am interested in arrangements of MATHEMATICS
with

both T's before both A's or
both A's before both M's or
both M's before the E

I define the following sets

$A_1$ = {arrangements with both T's before both A's}

$A_2$ = {arrangements with both A's before both M's}

$A_3$ = {arrangements with both M's before E}

I am able to find the cardinality of everything else but $A_1 \cap A_3$

i.e. arrangements with both T's before both A's and both M's before E.

I know I want to pick four of the 11 spots (11 choose 4)

and arrange H,I,C,S in those spots (4! ways)

But I dont know whaat to do next, I can't place only the T's in the two left most spots because certainly go T,M,M,E,T.

2. ## Re: Counting using inclusion/exclusion principle.

Originally Posted by Jame
I am interested in arrangements of MATHEMATICS with
both T's before both A's or
both A's before both M's or
both M's before the E
I define the following sets
$A_1$ = {arrangements with both T's before both A's}
$A_2$ = {arrangements with both A's before both M's}
$A_3$ = {arrangements with both M's before E}
I am able to find the cardinality of everything else but $A_1 \cap A_3$
Surely $\|A_1 \cap A_3\|=\binom{11}{4}(4!)\binom{7}{4}$.

i.e. Place the $\{H,I,C,S\}$ and arrange; select the four places to place $\{T,T,A,A\}$(only one way); finally put the $\{M,M,E\}$ is the remaining three places.