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# Ida had 5 cards with matching envelopes - same in design, different in

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Manager
Joined: 17 May 2018
Posts: 139
Location: India
Re: Ida had 5 cards with matching envelopes - same in design, different in  [#permalink]

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15 Apr 2019, 22:06
sayan640 wrote:
sumi747
Did you find it difficult to understand anywhere ?

Posted from my mobile device

I got the derangement solution just wanted a different approach without the use of a new formula
GMAT Club Legend
Joined: 18 Aug 2017
Posts: 6436
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: Ida had 5 cards with matching envelopes - same in design, different in  [#permalink]

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15 Apr 2019, 23:29
sumi747 wrote:
sayan640 wrote:
sumi747
Did you find it difficult to understand anywhere ?

Posted from my mobile device

I got the derangement solution just wanted a different approach without the use of a new formula

sumi747 ; see below

Total no of ways = 5! = 120

my take on question below as per my understanding of the link
total cards given 5

so chances of selecting 1 correct card into 1 correct envelope = 5c1
total ways of choosing a card for an envelope = 5!
now we have to select ways to select a card which is to put into wrong envelope
so for 4 cards we have 3 ways to choose wrong envelope
for 3 cards we have 3 ways to choose wrong envelope
and for 2 cards we have 1 way to choose wrong envelope
total ways we can put 4 cards in wrong envelope; 3*3*1 = 9
The probability that exactly one card got into the matching envelope
5c1*9/5! = 3/8
Manager
Joined: 15 Nov 2017
Posts: 50
Re: Ida had 5 cards with matching envelopes - same in design, different in  [#permalink]

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16 Apr 2019, 06:03
Hi Bunuel,

Thank you!

KHow
Bunuel wrote:
doomedcat wrote:
ida had 5 cards with matching envelopes - same in design, different in color. She removed the cards from all envelopes and randomly put them back. What is the probability that exactly one card got into the matching envelope?

A $$\frac{3}{16}$$
B $$\frac{5}{8}$$
C $$\frac{3}{8}$$
D $$\frac{1}{4}$$
E $$\frac{1}{2}$$

Source : Experts Global

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Manager
Joined: 05 May 2019
Posts: 150
Re: Ida had 5 cards with matching envelopes - same in design, different in  [#permalink]

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01 Jun 2020, 21:56
sayan640 wrote:

Total no of ways = 5! = 120

No of ways one card can be chosen which goes into the correct envelope o

ut of 5 cards is = 5 c 1 = 5

The no of ways in which other four cards can be put into wrong envelopes = D (4) = 9
So the required probability = 5 * 9 / 120 = 45 / 120 = 3 /8

I didn't really get
The no of ways in which other four cards can be put into wrong envelopes = D (4) = 9
Re: Ida had 5 cards with matching envelopes - same in design, different in   [#permalink] 01 Jun 2020, 21:56

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