Permutations and Combinations question (1 Viewer)

[Hermes1]

How many arrangements can be made by the letters of the word DEFINITION:

if the letter I does not occupy the first or last place.

 

[Hermes1]

I managed to solve it couple of minutes after i posted but i will leave this up 4 anyone who's interested.

 

[someth1ng]

Normal Arrangements: 10!/3!2!=302400
First+Last is I: 8!/2!=20160
I First Only: 9!/2!2! - 8!/2!=70560
I Last Only: 9!/2!2! - 8!/2!=70560

.`. Arrangements with no I in first or last is: 302400-20160-70560-70560=141120 ways

Is this correct? It looks okay to me but I'm not sure, could be a mistake in it.

 

[thoth1]

yeh that is correct.

 

[funnytomato]

I First Only: 9!/2!2! - 8!/2!=70560
I Last Only: 9!/2!2! - 8!/2!=70560

?

 

[meilz_is_a_slut]

That's bout the easiest perm/comb question you will ever get. Come on, you do 4unit, stop failing!

 

[Hermes1]

I First Only: 9!/2!2! - 8!/2!=70560
I Last Only: 9!/2!2! - 8!/2!=70560

?

i didnt use that guys way either lol but he got the right answer.

 

[funnytomato]

sorry i didn't see the subtracting 8!/2! part

 

[someth1ng]

i didnt use that guys way either lol but he got the right answer.

Would I be able to see your method aswell? I'd actually want to know if there was a simpler way.
Cheers,

 

[Hermes1]

Would I be able to see your method aswell? I'd actually want to know if there was a simpler way.
Cheers,

i dont think my way is quicker. i have just started perms and combs at skool so i dont fully no all the tricks yet. but heres wat i did:
i used 3 cases:
case 1: N is both at front and back
1x(8!/3!)
case 2: N occupies one place either at front or back
2P1x5P1x(8!/3!)
=10x(8!/3!)
case 3: N occupies neither front or back
5P1x4P1x(8!/3!2!)
=20x(8!/3!2)
=10x(8!/3!)

therefore total = 21x(8!/3!) = 141,120.

i dont no if this is quicker but this is how i did it.

 

[kooliskool]

There's actually a faster way.

If you think that there are 7 letters that can be on either side, disregarding the 3 I's (with order), you do that first: 7P2

Then arrange the rest of the 8 letters: 8!

But because of repetition of N and I, divide 2! and 3!

Hence the answer is 7P2×8!÷2!÷3!=141120

 

[Hermes1]

ah ok.

 

[someth1ng]

There's actually a faster way.

If you think that there are 7 letters that can be on either side, disregarding the 3 I's (with order), you do that first: 7P2

Then arrange the rest of the 8 letters: 8!

But because of repetition of N and I, divide 2! and 3!

Hence the answer is 7P2×8!÷2!÷3!=141120

That is a faster way but I don't think I'd be able to think of that method in the exam.
Hmm...so it's 7P2 for the first/last where 2xN is compensated at the end.
8! for the middle where repeats are compensated at the end.

It's a really short method but I personally think it requires abit more thought XD