Thread: Permutation Problem

1. Permutation Problem

Problem: In how many ways can these letters EFGHIEFGHI be arranged if the same letter musn't appear next to each other?

since we have 2 E's, F's, G's, H's and I's... The formula should be

10!
------------- = 113400 ways
2!2!2!2!2!

Then I combine similar letters [EE]FGHIFGHI for 5 letters

9!x5
-------- = 113400 ways
2!2!2!2!

That's where i get stumped... need help... XD

2. Originally Posted by mcroldan08
Problem: In how many ways can these letters EFGHIEFGHI be arranged if the same letter musn't appear next to each other?
You have started it correctly. Now we apply the inclusion/exclusion rule.
$\sum\limits_{k = 0}^5 {\left( { - 1} \right)^k \binom{5}{k}\frac{{\left( {10 - k}\right)!}}{{2^{5 - k} }}}$

3. uhmm... could you walk me through that rule... I'm not familiar with it... XD

4. Originally Posted by mcroldan08
uhmm... could you walk me through that rule... I'm not familiar with it... XD
If that is true, then I must wonder why you are asked to do this problem?
Inclusion-Exclusion Principle -- from Wolfram MathWorld

5. =n(A) + n(B) + n(C) + n(D) + n(E) - n (A&B) - n(A&C) - n(A&D) - ... - n(A&B&C) - n(B&C&D) - ... - n(A&B&C&D) - n(A&B&C&E) - ... + n(A and B and C and D and E)

Is this right?? XD

6. Originally Posted by mcroldan08
=n(A) + n(B) + n(C) + n(D) + n(E) - n (A&B) - n(A&C) - n(A&D) - ... - n(A&B&C) - n(B&C&D) - ... - n(A&B&C&D) - n(A&B&C&E) - ... + n(A and B and C and D and E)

Is this right?? XD
That is the basic idea.

7. Got it...

Final Check...

= 5x9!/2!2!2!2! - 10x(8!/2!2!2!) - 10x(7!/2!2!) - 5x(6!/2!) + 5!
=48720

so 113400-48720 = 64680...

anything I messed up??

btw... A heap of thanks!! really really appreciated the help...

8. Originally Posted by mcroldan08
Got it...

Final Check...

= 5x9!/2!2!2!2! - 10x(8!/2!2!2!) - 10x(7!/2!2!) - 5x(6!/2!) + 5!

anything I messed up??
You missed a sign. It should be +.

9. Ok.. Got it!! Thanks a lot!!