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

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Re: Ida had 5 cards with matching envelopes - same in design, different in  [#permalink]

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New post 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
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Re: Ida had 5 cards with matching envelopes - same in design, different in  [#permalink]

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New post 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
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Re: Ida had 5 cards with matching envelopes - same in design, different in  [#permalink]

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New post 16 Apr 2019, 06:03
Hi Bunuel,

Could you please provide your reasoning for how to solve this?

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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Hope it helps.
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Re: Ida had 5 cards with matching envelopes - same in design, different in  [#permalink]

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New post 01 Jun 2020, 21:56
sayan640 wrote:
Bunuel , VeritasKarishma, mikemcgarry

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
Can you please explain?
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Re: Ida had 5 cards with matching envelopes - same in design, different in   [#permalink] 01 Jun 2020, 21:56

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