Problem Solving

GmatKiss wrote:Alicia purchases three different rings that can each be worn on any of her fingers, excluding her thumbs. If she wants to wear at least one ring on each hand, with no more than one ring per finger, how many different ways can she distribute the rings among her eight fingers?

A. 192
B. 288
C. 336
D. 415
E. 465

Here's one approach.

Let's first ignore the rule about " at least one ring on each hand"

We'll take the task of placing the 3 rings on different fingers and break it into stages.

Stage 1: Place the first ring on a finger: There are 8 fingers, so this stage can be accomplished in 8 ways.
Stage 2: Place the second ring on a finger: There are 7 fingers remaining, so this stage can be accomplished in 7 ways.
Stage 3: Place the third ring on a finger: There are 6 fingers remaining, so this stage can be accomplished in 6 ways.

When we apply the Fundamental Counting Principle, we see that the number of ways to accomplish all 3 stages (and place the 3 rings) = 8x7x6 = 336

Of course among these 336 arrangements, there are some that break the rule about having at least one ring on each hand. In other words, we have counted arrangements where there are zero fingers on a hand. We need to count these arrangements and subtract them from 336. There are 2 cases to consider:
- case a) zero rings on the left hand
- case b) zero rings on the right hand

case a: This means that all 3 rings are on the right hand.
In how many ways can we place all 3 rings on the right hand?
Stage 1: place the first ring on a finger (4 ways)
Stage 2: place the second ring on a finger (3 ways)
Stage 3: place the third ring on a finger (2 ways)
Total = 4x3x2 = 24

case b: This means that all 3 rings are on the left hand.
Following the same steps as in case a, we get:
Total = 4x3x2 = 24


So, total number of ways to wear all three rings such that there is at least one ring on each hand = 336 - 24 - 24 = 288 = B

Cheers,
Brent

This is what I love about GMAT math questions! There can be several different approaches that all work.

Cheers,
Brent

coolhabhi wrote:

Brent@GMATPrepNow wrote:This is what I love about GMAT math questions! There can be several different approaches that all work.

Cheers,
Brent

Brent I have a doubt. I have done it this way.

2 rings on 1 hand and 1 ring on another hand. So
Choosing 1 hand = 2C1 = 2
1st ring on that hand = 4 (Since fingers =4) and
2nd ring on that hand = 3 (Since fingers =3)
3rd ring on another hand = 4
So it will be 2*4*3*4 = 96

Similarly the other hand = 96.

Total ways = 96 + 96 = 192. What am I missing. Please help me..

Your solution assumes that ring #1 and ring #2 must go on the same hand.

Cheers,
Brent