Basic P&C

[parveen110]

Hi,

There are five different caps and five different boxes. The capacity of each box is sufficient to accomodate all the five caps.

Q.1. If all the caps are of different colors and each box can have only one cap, in how many ways can you arrange the caps among the five boxes?

Q.2. If atleast one cap has to be distributed and the caps have to be arranged such that any box can have a maximum of one cap only, in how many ways can you arrange the caps among the five boxes?

I am not able to distinguish between the two questions, that probably has come from weak understanding of the concepts. Kindly post the conceptual difference between the two. Thank you.

P.S.: I am deliberately not posting the answer options.

Thanks!!

[Patrick_GMATFix]

Q1. Rephrase: how many ways can we place 5 caps in 5 boxes, if each box has 1 cap. Suppose the boxes represent positions; we can think of this question as how many ways can we order 5 caps? This is simply 5! = 120 ways.

Q2. "at least one cap has to be distributed" suggests that we could arrange all 5 caps as in Q1, but we could also use just 4 caps, or just 3, 2, or 1 cap.

If we find out how many ways we can place 1 cap, then 2, 3 and 4 caps, we can add up all those ways to 120 (ways to place 5 caps) to find out all possible outcomes.

Each of these ways can be broken into 3 steps.
(1) select the caps,
(2) select the boxes that will contain the selected caps,
(3) find # of ways to arrange the selected caps into the selected boxes

Multiplying (1) x (2) x (3) will give us all the ways to place the selected number of caps into boxes

Let's try this out

Ways to place 1 cap in 5 boxes.
(1) there are 5 ways to pick 1 cap from 5.
(2) there are 5 ways to pick 1 box from 5.
(3) there is only 1 way to arrange 1 cap in 1 box.
So there are 5 * 5 * 1 = 25 ways to use 1 cap. This should make conceptual sense since there are 5 ways we can put a cap in a box, and 5 caps we can pick from.

Ways to place 5 caps in 5 boxes.
(1) there is 1 way to pick 5 caps from 5.
(2) there is 1 way to pick 5 boxes from 5 to contain the caps
(3) there are 5! = 120 ways to arrange 5 caps in 5 boxes.
So there are 1 * 1 * 120 = 120 ways to use 5 caps. This is consistent with what we found from Q1 above.

Ways to place 2 caps in 5 boxes.
(1) there are 5!/(2!3!) = 10 ways to pick 2 caps.
(2) there are 10 ways to pick 2 boxes to contain the caps
(3) there are 2! = 2 ways to arrange the 2 caps in the 2 boxes (1st cap in 1st box, 2nd cap in 2nd box or reverse)
So there are 10 * 10 * 2 = 200 ways to use 2 caps.

Ways to place 3 caps in 5 boxes.
(1) there are 10 ways to pick 3 caps
(2) there are 10 ways to pick 3 boxes to contain the caps
(3) there are 3! = 6 ways to arrange the 3 caps in the 3 boxes.
So there are 10 * 10 * 6 = 600 ways to use 3 caps.

Ways to place 4 caps in 5 boxes.
(1) there are 5!/(4!1!) = 5 ways to pick 4 caps (makes sense; 5 ways to choose the cap that does NOT get picked)
(2) there are 5 ways to pick 4 boxes to contain the caps.
(3) there are 4! = 24 ways to arrange the 4 caps in the 4 boxes.
So there are 5 * 5 * 24 = 600 ways to use 4 caps.

So if we consider all possible selections (we can pick any number of caps from 1 to 6), there are 25 + 120 + 200 + 600 + 600 = 1545 possible outcomes