Permutations question - help! (1 Viewer)

[lisraeli]

Hey guys, I'm finding these types of questions kind've difficult. These two questions are similar I think you just need to know the approach.

1.how many 4 letter arrangements are possible from the word
DIVERSITY

2. How many different 4 letter arrangements are possible using the 12 letters from

SLEEPY SENATE (question taken out of Gaha's book, I didnt understand his solution)

Thanks a lot!

 

[Riviet]

1. Use three cases just like Geha's illustrated on page 175

2. The five cases in his solution are all the possible letter combinations that you can get with a 4 letter word using the letters in SLEEPY SENATE. Hence by counting the number of ways in each case and adding them up, you get the total number of 4 letter arrangements.

 

[lisraeli]

In his "Tip 41", I think hes made a mistake. I think hes considered SYSTEM to have 7 letters instead of 6 because otherwise i dont think it makes sense.

Anyway, would DIVERSITY be:

Case 1: 2 i's

= (2c2 . 7c2 . 4!)/2! = 252

Case 2: 1 i

= 2c1 . 7c3 . 4! = 1680

Case 3 : no i's

= 7c4 x 4! = 840

252 + 840 + 1680 = 2772

??

I still dont really understand the solution for SLEEPY SENATE

 

[Riviet]

Yep, it's an error for the SYSTEM question, should be 6 letters. Your solution for DIVERSITY looks right to me.

 

[adosh]

<HR style="COLOR: #d1d1e1" SIZE=1><!-- / icon and title --><!-- message -->hello everyone,,, im having a problem wtih permuts combins,,,,now geha answers a questions as such "in how many ways can 8 people be arranged into two sets of four for tennis??" : (8C4x4C4)/2!= 35,,,now i understand this but heres a similar fitzpatrick question

find the number of wats in which 6 women and 6 men can be arranged in three sets of four for tennis wihcout restrictions??? and the asnwer is 155925.....now why cant we apply gehas method to this questuion, that is

(12C4x8C4x4C4)/3!=5775 but this is way off from 155925,,,,couls someone please help..thanks in advance

 

[bored of sc]

Find the number of ways in which 6 women and 6 men can be arranged in three sets of four for tennis without restrictions?

12C4.8C4.4C4 = 34650 <--- number of selections for sets of 4

4! = 24 <--- number of arrangements for each set of people

That's as far as I can get.


Maybe you have to treat the men and women like they are the same:
(6C4.6C0*2+6C3.6C1*2+6C2.6C2) = 495 for the first set.

etc

 

[u-borat]

this approach should work:

there are 12! ways of arranging 12 people.
divide it by 2! for every pair of people (because AB is the same as BA)
divide it by 2! for every court side (AB playing CD is the same as CD playing AB)
divide it by 3! because you have 3 games and there is no order for the games being played.

so its 12!/ 9(2!) x 3! = 155925

 

[bored of sc]

12!/ 9(2!) x 3!

That expression doesn't equal 155925

 

[u-borat]

ye sorry it should read 2!^9

 

[bored of sc]

ye sorry it should read 2!^9

Oh yeah, of course 2!*2! ... etc = 2!⁹.