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GmatMathPro Problem

Expert replies
by kakz » Sat Oct 01, 2011 5:20 am
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

oa is b.
I got this one in the gmatmathpro.com site. Hope it dint' mind me posting here.
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Source: — Problem Solving |

by shankar.ashwin » Sat Oct 01, 2011 5:45 am
2 cases;

First 2 ring in one hand and third in the other

8*3*4 = 96

First ring in one hand, second in the other and third anywhere

8*4*6 = 192

Total Cases = 192+96=288
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by GMATGuruNY » Sat Oct 01, 2011 5:46 am
kakz 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

oa is b.
I got this one in the gmatmathpro.com site. Hope it dint' mind me posting here.
A pair of fingers on one of the hands must bear rings:
Number of ways to choose a pair from 4 fingers = 4C2 = 6.
Since this pair of fingers could be on either of the 2 hands, we multiply by 2:
6*2 = 12.

Number of ways to ARRANGE 2 of the 3 rings on this pair of fingers:
3*2 = 6.

3rd ring:
This ring must be worn on one of the 4 fingers on the OTHER hand.
Number of options = 4.

To combine the options above, we multiply:
12*6*4 = 288.

The correct answer is B.
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by Brent@GMATPrepNow » Sat Oct 01, 2011 5:57 am
kakz 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

oa is b.
I got this one in the gmatmathpro.com site. Hope it dint' mind me posting here.
The great thing about counting questions is that there is often more than 1 approach. Here's another option:

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
Brent Hanneson - Creator of GMATPrepNow.com
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