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

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Math Expert
Joined: 02 Sep 2009
Posts: 61301
Alicia purchases three different rings that can each be worn on any of  [#permalink]

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27 Nov 2019, 00:47
00:00

Difficulty:

65% (hard)

Question Stats:

49% (02:12) correct 51% (02:25) wrong based on 35 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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Joined: 19 Oct 2018
Posts: 1301
Location: India
Re: Alicia purchases three different rings that can each be worn on any of  [#permalink]

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27 Nov 2019, 04:26
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
Math Expert
Joined: 02 Aug 2009
Posts: 8256
Re: Alicia purchases three different rings that can each be worn on any of  [#permalink]

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27 Nov 2019, 05:17
1
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

B
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Joined: 18 Aug 2017
Posts: 5931
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: Alicia purchases three different rings that can each be worn on any of  [#permalink]

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27 Nov 2019, 09:46
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
IMO B
Re: Alicia purchases three different rings that can each be worn on any of   [#permalink] 27 Nov 2019, 09:46
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