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

Selecting one right envelope = 5 C 1 = 5 ways
Derangment of 4 envelopes = 4!(1- 1/1!+1/2!-1/3!+1/4!) = 9
Total ways of arrangement = 120

Probability of 1 envelope placed in right place = 9*5/120 = 3/8

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Total no of ways = 5! = 120

No of ways one card can be chosen which goes into the correct envelope out 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[/quote]


This is correct.
Not sure what the official explanation says and why the official answer is (1/4) if it is indeed 1/4.[/quote]

Bunuel
Please correct the OA...Unnecessarily giving way to confusion....Please change the OA from D to option C...You may refer to my explanation for OE... [/quote]


VeritasKarishma , sayan640
hi could you please advise on the highlighted part. How did we arrive at 9?

akdebonair well I have never come across this formula in probability of de arrangement , anyways thanks for the explanation

Bunuel , VeritasKarishma, mikemcgarry

Total no of ways = 5! = 120

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

total cards = 5
right ; 1/5
wrong ; 3/4*2/3*1/2 *1/1
total
1/5*3/4*2/3*1/2 *1/1 * 5c4 ways = 1/4
IMO D

This is correct.
Not sure what the official explanation says and why the official answer is (1/4) if it is indeed 1/4.

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

=> Cards : \(C_1\), \(C_2\),\(C_3\),\(C_4\),\(C_5\).

=> Envelopes: \(E_1\), \(E_2\),\(E_3\),\(E_4\),\(E_5\).(Same design, different color).

The probability that exactly one card got into the matching envelope

=> Probability : \(\frac{Desired results }{ Total number of results.}\)

5 cards can be put into 5! ways = 120.

Exact one card correct:

Example : \(C_1\) goes in \(E_1\) (Correct).

=> No of ways one card can be chosen which go into the correct envelope out of 5 cards is = \(^{5}\mathrm{C_1}\) = 5
= 5.

No of ways \(C_2\) goes into the wrong envelope: \(E_3\), \(E_4\), and \(E_5\) = 3 ways - (Suppose \(C_2\) has been put in \(E_3\)).

No of ways \(C_3\) goes into the wrong envelope: \(E_2\), \(E_4\), and \(E_5\) = 3 ways - (Suppose \(C_3\) has been put in \(E_2\)).

No of ways \(C_4\) goes into the wrong envelope: \(E_5\) = 1 way

No of ways \(C_5\) goes into the wrong envelope: \(E_4\) = 1 way

Total ways : 3 * 3 * 1 * 1 = 9

Desired result: 5 * 9

Total results: 120

Probability: \(\frac{5 * 9 }{ 120}\) = \(\frac{45 }{ 120}\) = \(\frac{3}{8}\)

Answer C

Bunuel
Please correct the OA...Unnecessarily giving way to confusion....Please change the OA from D to option C...You may refer to my explanation for OE...

________________________________
Done. Thank you.

It is formula for derangment derangment of n items is given by following formula.
D(n) = n!(1- 1/1!+1/2!-1/3!+1/4!......1/n!)

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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

This is correct.
Not sure what the official explanation says and why the official answer is (1/4) if it is indeed 1/4.[/quote]

Bunuel
Please correct the OA...Unnecessarily giving way to confusion....Please change the OA from D to option C...You may refer to my explanation for OE... [/quote]




VeritasKarishma , sayan640
hi could you please advise on the highlighted part. How did we arrive at 9?[/quote]

Hi Archit Archit3110,

Please go through this...

https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/blog/2011/ ... envelopes/

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

Total no of ways = 5! = 120

No of ways one card can be chosen which goes into the correct envelope out 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
( It has been derived in this link :- https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/blog/2011/ ... envelopes/ )
( In question 2 ..no 5 ..to make your life easier..)
Now I think you understood how D(4) = 9..

So the required probability = 5 * 9 / 120 = 45 / 120 = 3 /8

Yes understood where I went wrong.. thank you sayan640

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Other than derangement do we have any other method of solving this?

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sumi747
Did you find it difficult to understand anywhere ?

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