Question

Find how many arrangements there are possible from the word STUDIO

a. The letter S is not the first letter
b. The letter O is not the first or last letter

Answer 1

\[\Large \left[\begin{matrix}S &T&U&D&I&O\\?&?&?&?&?&?\end{matrix}\right]\]

Answer 2

You can think of it as counting the number of 'slots' each letter may go to.
\[\Large \left[\begin{matrix}S &T&U&D&I&O\\1&2&3&4&5&6\end{matrix}\right]\]

Answer 3

so 5!*6!?

Answer 4

That's way too much...
Even more than the number of rearrangements of STUDIO without restrictions...

Just hear me out first :P

Answer 5

sure

Answer 6

Now, let's consider the first restriction, namely, that S does not take the first position, or in other words, it may not take position#1

In that case, how many 'slots' are available for S?

Answer 7

5

Answer 8

That's right :)
Now, once S is placed, how many 'slots' are available for T?
There is no restriction for T, however, one of the slots is now currently occupied by S...

Answer 9

5

Answer 10

That makes 5*5 possibilities so far...
Catch me so far?

What about the letter U?

Answer 11

4

Answer 12

And it goes on, right? :)
5*5*4*3*2*1
or
\[\Large 5\cdot 5!\]

Answer 13

ok i got that

Answer 14

I'm sorry, there IS an easier way to do that...
First you count the number of rearrangements without restrictions, aye?
that's exactly 6!

Answer 15

yes

Answer 16

And then, you count precisely those rearrangements that DO have S on the first position, and that's like placing S on the first position and arranging the rest (there are 5 remaining)
and that's 5!

Answer 17

?

Answer 18

We count the ones that don't obey the restriction. Namely, the number of rearrangements that DO have S on the first position.
So that means, we fix S on the first position...
\[\Large \left[\begin{matrix}S &T&U&D&I&O\\S&?&?&?&?&?\end{matrix}\right]\]

Answer 19

And then rearrange the remaining letters on the remaining five slots, and that gives 5! ways, right?

Answer 20

I'll stop if you find this more confusing ^_^
Just let me know...

Answer 21

\[\Large 5\cdot 5!\]IS the answer for the first bit, I'm just showing you another way to get it.

Answer 22

i think i understood the first method better.

Answer 23

Yeah...it is more down-to-earth. Oh well :D Let's do the second?

Answer 24

For the next part i came up with 4.4! but its not the correct answer. i used the first method

Answer 25

Well, you can't just do these things methodically... :D

Answer 26

Here...
\[\Large \left[\begin{matrix}S &T&U&D&I&O\\1&2&3&4&5&6\end{matrix}\right]\]

Answer 27

there are a total of 4 positions for O

Answer 28

That's right.
So once O is placed, let's place the rest :D

Answer 29

But then again, the next letter will STILL have 5 possibilities for it, since only the letter O has any restrictions at all...

Answer 30

ok

Answer 31

Right? ^_^
That was your error.

Answer 32

so 4.5!?

Answer 33

mhmm :)
\[\Large 4\cdot 5!\]

Answer 34

i think my main concern is not understanding the concept well...

Answer 35

Combinatorics calls for concepts a lot.
Practice, practice, practice! :D

Answer 36

oh i've heard of that before eh?

Answer 37

thanks

Answer 38

Only naturally :D