Possible Arrangements

[BearMathTeam]

How many arrangements of five letters are possible using three letters from the word SQUARE and two letters from the word FOOT?

 

[pka]

How many arrangements of five letters are possible using three letters from the word SQUARE and two letters from the word FOOT?

The difficulty cones in with the possible use of the double \(\displaystyle OO\). Consider the collection \(\displaystyle <O,O,S,Q,R>\).
There are \(\displaystyle \dfrac{5!}{2!}\) ways to arrange those five letters. Why divide by \(\displaystyle \bf 2!~?\)
You must tell us from the letters \(\displaystyle <S,Q,U,A,R,E>~\&~<F,O,O,T>\) you can choose three from the first collection and two from the second.
Then tell us how many contain \(\displaystyle O,O\) and how many don't? How many ways to rearrange each kind?

 

[BearMathTeam]

The difficulty cones in with the possible use of the double \(\displaystyle OO\). Consider the collection \(\displaystyle <O,O,S,Q,R>\).
There are \(\displaystyle \dfrac{5!}{2!}\) ways to arrange those five letters. Why divide by \(\displaystyle \bf 2!~?\)
You must tell us from the letters \(\displaystyle <S,Q,U,A,R,E>~\&~<F,O,O,T>\) you can choose three from the first collection and two from the second.
Then tell us how many contain \(\displaystyle O,O\) and how many don't? How many ways to rearrange each kind?

I figured out the answer: \(\displaystyle {6 \choose 3} *\bf 5! ~ for <SQUFT>.....<AREFT>\) , Similarly \(\displaystyle {6 \choose 3}*\bf 5! ~ for <SQUFO>.....<AREFO>\) and
\(\displaystyle {6 \choose 3} *\bf 5! ~ for <SQUOT>.....<AREOT>\) and Now like you said, OO gets a different result: \(\displaystyle {6 \choose 3} *\dfrac{5!}{2!} ~ for <SQUOO>.....<AREOO>\) Hence we have 2400+2400+2400+1200 = 8400

 

[Steven G]

I figured out the answer: \(\displaystyle {6 \choose 3} *\bf 5! ~ for <SQUFT>.....<AREFT>\) , Similarly \(\displaystyle {6 \choose 3}*\bf 5! ~ for <SQUFO>.....<AREFO>\) and
\(\displaystyle {6 \choose 3} *\bf 5! ~ for <SQUOT>.....<AREOT>\) and Now like you said, OO gets a different result: \(\displaystyle {6 \choose 3} *\dfrac{5!}{2!} ~ for <SQUOO>.....<AREOO>\) Hence we have 2400+2400+2400+1200 = 8400

Suppose there were 4 distinct letters in the word foot. Then the answer to your problem would be 6C3*4C2=120. How can you get 8400 if your 2nd word has only three distinct letters??

 

[pka]

I figured out the answer: \(\displaystyle {6 \choose 3} *\bf 5! ~ for <SQUFT>.....<AREFT>\) , Similarly \(\displaystyle {6 \choose 3}*\bf 5! ~ for <SQUFO>.....<AREFO>\) and \(\displaystyle {6 \choose 3} *\bf 5! ~ for <SQUOT>.....<AREOT>\) and Now like you said, OO gets a different result: \(\displaystyle {6 \choose 3} *\dfrac{5!}{2!} ~ for <SQUOO>.....<AREOO>\) Hence we have 2400+2400+2400+1200 = 8400

Congratulations your answer is correct.
Here is just a comment on a shorter method:
Choosing three letters from \(\displaystyle \mathit{SQUARE}\) and two from \(\displaystyle \mathit{FOOT}\).
There are \(\displaystyle \dbinom{6}{3}=20\) ways to choose the three.
The other letters must be one of these \(\displaystyle <\mathit{F,T}>,<\mathit{F,O}>,<\mathit{O,T}>,<\mathit{O,O}>\)
The first three of those gives us \(\displaystyle 60\) choices five distinct letters.
But using \(\displaystyle <\mathit{O,O}>\) get twenty selections of five containing two identical.
So \(\displaystyle 60\cdot 5!+20\cdot\frac{5!}{2}=8400\). SEE HERE