pizza

OA [spoiler]= E: 260[/spoiler] ?

5*4*4 (assume: two quadrants with the same topping)

5*4*3*3 (assume: All quadrants with different toppings)

Sum [spoiler]= 260[/spoiler]

Correct me if I am wrong !

Sorry I missed the OA. It's 70

What's wrong with this?
First quadrant: 5C1
Second: 4C1
Third: 4C1
Fourth: 3C1
Total ways = (5C1*4C1*4C1*3C1) = 240

sunilrawat wrote:A circular pizza is cut into 4 quadrants. Each piece can be topped with any one of five available toppings but no two adjacent pieces have the same topping. In how many ways can the pieces be topped?
40
70
160
240
260

oops! my bad

great work shankar.ashwin

Tricky one, this one is also a circular arrangement. So we need to keep that in mind.

4 different toppings
5C4 - 5 ways of picking up toppings and (4-1)! ways of arranging them. 5*6 = 30

3 Different toppings
5C3 = 10 ways of picking up toppings and 3 ways to arrange them. 10*3 = 30

2 different toppings

5C2 - 10. and only 1 way to arrange.

Together 30+30+10 = 70

sunilrawat wrote:Sorry I missed the OA. It's 70

What's wrong with this?
First quadrant: 5C1
Second: 4C1
Third: 4C1
Fourth: 3C1
Total ways = (5C1*4C1*4C1*3C1) = 240

Besides the fact that it doesn't take into account the circular permutations, at least one thing that's wrong is that what you can choose for the fourth quadrant depends on what was chosen for the third quadrant. For example, let's say we have toppings A,B,C,D,E. You have 5 choices for the first quadrant, so let's say you pick A. Then you have 4 choices for the 2nd quadrant, so let's say you pick B. Now, for the third quadrant you can pick from A,C,D or E. But of you pick A, then the 4th quadrant would have 4 choices; it could be anything besides A. If you pick C, D, or E, then the 4th quadrant could have only 3 choices. Thus, it won't always be the case that you have 3 choices for the 4th quadrant.

shankar's way looks good to me, though I would probably change 5*3=30 to 10*3=30 in the second case.

Typo and edited

GmatMathPro wrote:
shankar's way looks good to me, though I would probably change 5*3=30 to 10*3=30 in the second case.

shankar.ashwin wrote:Tricky one, this one is also a circular arrangement. So we need to keep that in mind.

4 different toppings
5C4 - 5 ways of picking up toppings and (4-1)! ways of arranging them. 5*6 = 30

3 Different toppings
5C3 = 10 ways of picking up toppings and 3 ways to arrange them. 10*3 = 30

2 different toppings

5C2 - 10. and only 1 way to arrange.

Together 30+30+10 = 70

Hi Shankar....

Can you plz explain how did u get 3 ways to arrange the three diff toppings.....

Hi, say you have picked up 3 toppings A,B and C. They can be arranged in the following ways.

2A's opposite to each other B&C on either side.
2B's opposite to each other A&C on either side.
2C's opposite to each other A&B on either side.

You can pick up 3 toppings from 5 in 5C3 ways and arrange each of them in 3 ways as above. So you get 10*3=30 ways

leonswati wrote:
Hi Shankar....

Can you plz explain how did u get 3 ways to arrange the three diff toppings.....

@ Shankar and Pete
Thanks.

Source: I googled for another permutation question and found this one in a forum, I think gmatclub.
There was OA without any explanation.