Permutations and combinatiosn help (2 Viewers)

[messagebrd]

Permutations and combinations help

Alrite guys i've been trying to do this question for a while now and I'm about to pull my hair out doing it.

It's question 31) of exercise 10E from the Cambridge 3u Year 12 Book.

Eight people are to form two queues of four. In how many ways can this be done if:
a) there are no restrictions = 8P4 x 4P4
b) Jim Will only stand on the left hand queue = 7P3 x 4P4 x 4
c) Sean and Liam must stand in the same queue: ??

Answer says: 17,280

 

[ascentyx]

The number of combinations is 2 x 4p2 x 6p6.

Pretty much 2 is from the fact that they can be in either of the two rows, the 4p2 is from the number of ways you can rearrange the two people in a row of 4, and 6p6 is the number of ways you can choose the remaining 6 others. Makes sense?

 

[h3ll h0und]

shudnt (b) be 7P3 x 4P4 ?

 

[messagebrd]

shudnt (b) be 7P3 x 4P4 ?

Why??

EDIT: I see where your coming from but that gives you 5040... i think then it's actually 7P3 x 4P4 x 4 (as he can move around in his queue in 4 different ways)

 

[messagebrd]

The number of combinations is 2 x 4p2 x 6p6.

Pretty much 2 is from the fact that they can be in either of the two rows, the 4p2 is from the number of ways you can rearrange the two people in a row of 4, and 6p6 is the number of ways you can choose the remaining 6 others. Makes sense?

Yeah i get it.. thanks a million

 

[h3ll h0und]

Why??

EDIT: I see where your coming from but that gives you 5040... i think then it's actually 7P3 x 4P4 x 4 (as he can move around in his queue in 4 different ways)

oh yea... lol missd that ... im noob at permutations and combinations lol >< wats the answer?

 

[messagebrd]

^ 20, 160

...Perms and combs has to be one of the most annoying 3u topics

 

[ascentyx]

Why??

EDIT: I see where your coming from but that gives you 5040... i think then it's actually 7P3 x 4P4 x 4 (as he can move around in his queue in 4 different ways)

b) would just be the same thing as c) but without the 2 x. Pretty much it's 4p2 (number of ways you can rearrange 2 people in 1 queue) x 6p6 (the number of ways the rest of the people can be rearranged).

 

[h3ll h0und]

^ 20, 160

...Perms and combs has to be one of the most annoying 3u topics

i noe! :burn: theres always somehting i miss in the question

 

[Drongoski]

Re: Permutations and combinations help

c) = 2 x { ₆P₂ x 2! x 3! x ₄P₄ } = 17280

 

[Drongoski]

^ 20, 160

...Perms and combs has to be one of the most annoying 3u topics

Agree with u entirely. As I've indicated elsewhere earlier, unless you have a very lucid mind, combinatorics can easily make a fool of you. So easy to overlook something here or there or to formulate the solution incorrectly. I dread it myself.