How many ways to arrange these portraits?

P

pgfreak

New member
Joined
Oct 15, 2006
Messages
9
Question:
There are 12 different portraits.
a) In how many ways can 5 of them be hung on the wall?
b) If picture A must be included and hung in the middle of the group, how many different arrangements are there?

Ok, so for a... I started with 12 x 11 x 10 x 9 x 8 to represent choosing the 12 portraits in the 5 spots (or 12p5).
But how do I account for the different places each one of the portraits can be placed (arranging)??? I am going to need to multiply what i just did (choosing) with the numbers of arrangements and I dont know how to find the arrangements, can anyone help me? (and explain)

(I didnt get to B yet because I want to do A first)
 
Hint: Think about the permutation formula they gave you! :wink:

Eliz.
 
nPr is what I know as the formula

or ... wouldnt there be 5! ways to arrange it? (5 factorial)

so 12 x 11 x 10 x 9 x 8 x 5! ?


EDIT: looking back... did I make this too complicated? Should it just be 12P5 ??

And then b would be 11p4?
 
\(\displaystyle P(N,j) = \frac{{N!}}{{\left( {N - j} \right)!}}\quad \Rightarrow \quad P(12,5) = \frac{{12!}}{{7!}} = \left( {12} \right)\left( {11} \right)\left( {10} \right)\left( 9 \right)\left( 8 \right)\)

Think of is as having 12 choices, 11 choices, 10 choices, etc.

For part b, we have _ _ A _ _. How may ways can we fill the blanks?
 
pka said:
\(\displaystyle P(N,j) = \frac{{N!}}{{\left( {N - j} \right)!}}\quad \Rightarrow \quad P(12,5) = \frac{{12!}}{{7!}} = \left( {12} \right)\left( {11} \right)\left( {10} \right)\left( 9 \right)\left( 8 \right)\)

Think of is as having 12 choices, 11 choices, 10 choices, etc.

For part b, we have _ _ A _ _. How may ways can we fill the blanks?
Click to expand...

SO my part A was correct then right? 12P5?

and b is 11p4 right? (11! / 7! )
 
You must log in or register to reply here.
Top