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Alicia purchases three different rings that can each be worn on any of

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Alicia purchases three different rings that can each be worn on any of  [#permalink]

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New post 27 Nov 2019, 01:47
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Question Stats:

38% (01:32) correct 62% (02:25) wrong based on 29 sessions

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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

Are You Up For the Challenge: 700 Level Questions

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Re: Alicia purchases three different rings that can each be worn on any of  [#permalink]

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New post 27 Nov 2019, 05:26
Choose 2 fingers of left hand and one finger of the right one or vice-versa. Total number of possible arrangements of 3 different rings = 3!

Total number of ways
= 2*4C2*4C1*3!
= 288


Bunuel 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

Are You Up For the Challenge: 700 Level Questions
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Re: Alicia purchases three different rings that can each be worn on any of  [#permalink]

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New post 27 Nov 2019, 06:17
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Bunuel 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

Are You Up For the Challenge: 700 Level Questions



Two ways.

(I) As also shown above
Choose 2 fingers on one hand and one finger on second hand =2*4C2*4C1=2*4*4*3/2=48...multiplication by 2 is because we can choose any hand for 2 fingers
Now these rings can be arranged in 3! Or 6 ways
Total = 48*6=288

(II) Total ways =8*7*6 ( first finger any of 8 fingers and so on)
Restricted ways = All three on same hand = 4*3*2*(2)
Total ways =8*7*6-4*3*2*2=336-48=288

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Re: Alicia purchases three different rings that can each be worn on any of  [#permalink]

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New post 27 Nov 2019, 10:46
Bunuel 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

Are You Up For the Challenge: 700 Level Questions


total ways to wear 3 different rings on ; 3!
and choosing 3 fingers among two hands ; 2c1 * 4c2* 4c1 ;
total possible ways ; 3!*2* 4c2*4c1 ; 288
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Re: Alicia purchases three different rings that can each be worn on any of  [#permalink]

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New post 02 Dec 2019, 07:50
Hi, Shouldn't the total number of possible arrangements be 4 * 3 * 2 = 24
and hence, it should be 4C1 * 4C2 * 2 * 24
Why is it 3! ?
Could someone please explain me this ?
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Re: Alicia purchases three different rings that can each be worn on any of   [#permalink] 02 Dec 2019, 07:50
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Alicia purchases three different rings that can each be worn on any of

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