Using the inclusion- Exclusion formula, compute how many arrangements of the 26 alphabet letters there are that contain none of the words MATH, RUNS, FROM?
Could someone please explain how to do this?
Solution says; 26! -3*23! +2*20!
Using the inclusion- Exclusion formula, compute how many arrangements of the 26 alphabet letters there are that contain none of the words MATH, RUNS, FROM?
Could someone please explain how to do this?
Solution says; 26! -3*23! +2*20!
It's hard to know what kind of explanation you will understand since you don't tell us what you do understand.
Do you understand that there are n! ways to arrange n distinct things? There are 26! ways to arrange the 26 letters in the alphabet.
That's the reason for that first "26!". We then subtract off the number of "forbidden" arrangements.
If an arrangement contains "MATH" then the other 26- 4= 22 letters can be in any order. There are 22! such arrangements. Further, the "MATH" can appear in any of 23 places-before or after the other letters or in the 21 places between the 22 letters. That gives 23(22!) arrangements that contain "MATH".
If an arrangement contains "RUNS", the same argument gives 23(22!) arrangements that contain "RUNS".
If an arrangement contains "FOR", the other 26- 3= 23 letters can appear in 23! arrangements and "FOR" can appear in any of 24 places among those 23 letters. There are 24(23!) arrangements that contain "FOR".
We need to subtract those from the 26!
BUT we those 23(22!) arrangements that contain "MATH" may include some that also contain "RUNS" or may include "FOR", or both. Those will get subtracted more than once. We need to calculate how many of those arrangements there are and add them back in.
The other two replies are solutions. Here is my solution.
Consider these three blocks: $\boxed{\text{MATH}},~\boxed{\text{RUNS}},~\& ~ \boxed{ \text {FROM}}$
To use inclusion/exclusion to find the number at least one of the blocks occurs we must find the ways one at a time; subtract ways two at a time; add back way in which all three occur. Each block is four letters long leaving 22 other letters. They with one block make 23 objects. Thus the three taken one at a time gives $3(23!)$
Now the fun begins. It is impossible the have blocks $\boxed{\text{FROM}}~\&~\boxed{\text{RUNS}}$ occur together. Right?
But the pair $\boxed{\text{MATH}}~\&~\boxed{\text{RUNS}}$ can occur together, no problem: $(26-8)+2=20)$ So $20! $
The pair $\boxed{\text{MATH}}~\&~\boxed{\text{FROM}}$ can occur together, only as $\boxed{\text{FROMATH}}$ : $(26-7)+1=20)$ So $20! $
Thus taken two at a time is $2(20!)$ And we arrive with at least one at a time in $3(23!)-2(20!)$ ways.
Subtract that from the total: $(26!)-3(23!)+2(20!)$.