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Ram had to post 5 letters to 5 different persons. He put the l

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Ram had to post 5 letters to 5 different persons. He put the l  [#permalink]

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New post 07 Dec 2019, 19:10
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GMATBusters’ Quant Quiz Question -6


Ram had to post 5 letters to 5 different persons. He put the letters in the address stamped envelop randomly. What is the probability that exactly two letters are sent to correct address?

A. 1/12
B. 1/6
C. 1/4
D. 1/3
E. 1/2

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Re: Ram had to post 5 letters to 5 different persons. He put the l  [#permalink]

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New post 07 Dec 2019, 21:08
Answer: A

total number of ways =5!=120

total number where two are correct =5C2=10

P=10/120=1/12
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Re: Ram had to post 5 letters to 5 different persons. He put the l  [#permalink]

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New post 07 Dec 2019, 22:32
Total possible number of arrangements = 5! = 120

Ways of electing the two envelopes which are sent to correct address = 5 C 2 = 10 ways

Derangement General Formula, $$D(n) =n!*\sum_{k=0}^{n} \frac{-1^{k}}{k!} $$

Derangment of the remaing 3 envelopes = 3!(1- 1/1!+1/2!-1/3!) = 2

The probability that exactly two letters are sent to correct address = 10*2/120 = 1/6

Answer: 1/6

Correct option: B
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Re: Ram had to post 5 letters to 5 different persons. He put the l  [#permalink]

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New post 08 Dec 2019, 01:38
1
total ways for 5 letters ; 5! ; 120
and for exactly 2 ; 5*4 ; 20
P = 20/120 ; 1/6
IMO B

Ram had to post 5 letters to 5 different persons. He put the letters in the address stamped envelop randomly. What is the probability that exactly two letters are sent to correct address?

A. 1/12
B. 1/6
C. 1/4
D. 1/3
E. 1/2
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Re: Ram had to post 5 letters to 5 different persons. He put the l  [#permalink]

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New post 08 Dec 2019, 01:45
Ans:A
Letters : L1,L2,L3,L4,L5
Address: A1,A2,A3,A4,A5

Total combination: L1 can go in ant 5 * L2 can go in rest 4 *...
Hence, Total combination=5*4*3*2*1=5!

Total combination with correct address: [L1 and (any one of the rest) = 4] +[L2 and (any of the rest except L1) = 3+.....
Hence , Total combination with correct address =10
OR

Total combination with correct address=5c2
Hence Probablity=10/5!=1/12
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Re: Ram had to post 5 letters to 5 different persons. He put the l  [#permalink]

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New post 08 Dec 2019, 03:03
Probability that exactly two letters are sent to correct address is same as the probability that exactly three letters are not sent to correct address

denominator is 5!= 120

Numerator is 5C3*3!(1-1/1!+1/2!-1/3!)= 5C3*2 = 20

answer=1/6 (B)
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Re: Ram had to post 5 letters to 5 different persons. He put the l  [#permalink]

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New post 08 Dec 2019, 05:01
Ram had to post 5 letters to 5 different persons. He put the letters in the address stamped envelop randomly. What is the probability that exactly two letters are sent to correct address?

A. 1/12
B. 1/6
C. 1/4
D. 1/3
E. 1/2

Let R denote the letter sent to correct address and W denote the letter sent to correct address. We are trying to find the probability of 2R3W.
Probability = number of ways to get 2R3W/total number of ways

Imagine stuffing envelopes randomly. Ram can put any of 5 letters into the first envelope followed by any of the remaining 4 into second envelope, any of the remaining 3 into the next, either of the remaining 2 into the next, and has no choice to make on the last, or 5*4*3*2*1.

number of ways total is 5! = 120.

number of ways to get 2R3W: Ram could fill and post the first letter to right address (1 way), then put the second one also to the correct one and post it to right address, then put either of the 3 wrong remaining letters in the next (3 ways), then put a wrong letter in the next (2 way) and last one definitely to the wrong one. That's 1*1*2*1*1 = 2.

But since it doesn't have to be the first envelope that has the Right address, it could be any of the 5 envelopes, the total ways to get 2R3W is 5*2 = 10.
So, the probability will be 10/120 = 1/12.

Alternatively, it can be solved by visual approach.
let's consider first and second to be posted to correct address,
1st correct and 2nd correct and 3rd wrong and 4th wrong and 5th wrong
So, the probability = 1/5 * 1/4 * 2/3 * 1/2 * 1 = 1/60.
So, there will be 5 different cases, so 5 * 1/60 = 1/12.

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Re: Ram had to post 5 letters to 5 different persons. He put the l  [#permalink]

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New post 08 Dec 2019, 05:26
The answer is A: 1/12.

Explanation:
First, we find the total number of cases. This is easy, as there are 5 letterboxes, and 5 different persons will receive the letters. So, first letterbox, 5 persons can receive. Second letterbox, 4 persons can receive the letter. Third letterbox, 3 persons can receive the letter, and so forth.

So total number of cases = 5*4*3*2*1 = 120.

Now, given constraint, EXACTLY 2 letters have to be sent to the correct address. So, we can form the case as:
RRWWW, WRRWW, WWRRW and so forth. The exact formula would be: \( 5!/2!*3! \), as 2 Rights and 3 Wrongs are grouped. Solving it, we get it as 10.

So total number of valid cases = 10.

So, probability = number of valid cases/total number of cases = 10/120 = 1/12, which is the answer.
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Re: Ram had to post 5 letters to 5 different persons. He put the l  [#permalink]

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New post Updated on: 08 Dec 2019, 22:48
Let the 5 letters be L1, L2, L3, L4 and L5 and their correct envelopes be E1, E2, E3, E4 and E5

For favourable outcomes:

Select any two letters.
Two letters out of 5 can be selected in 5C2 ways.
The selected 2 letters can be put in their respective envelopes in only 1 way i.e. 5C2 × 1 = 5C2 ways. (e.g. L2 and L3 were selected, they can be put in E2 and E3 in only 1 way.)

But to get exactly 2 letters correctly posted, the remaining 3 letters must be put in the incorrect envelopes.
Let the remaining letters and envelopes be L1, L4, L5 and E1, E4, E5.

Correct combination is
L1 - E1
L4 - E4
L5 - E5

Possible incorrect combinations are
1. L1-E4, L4-E5, L5-E1.
2. L1-E5, L4-E1, L5-E4.

Therefore, total number of favourable outcomes will be
combination of correct letters * combination of incorrect letters = 5C2 * 2


For total number of outcomes:

L1 can be put in any of the 5 envelopes. Therefore, 4 envelopes will be left for 4 letters.

Now, L2 can be put in any of the remaining 4 envelopes. Therefore, 3 envelopes will be left for 3 letters.

Similarly, L3, L4 and L5 can be put in any of the remaining 3, 2 and 1 envelopes respectively.
Therefore, total number of outcomes = 5×4×3×2×1 = 5!

Therefore, probability, P = favouravle outcomes/ total no. of outcomes = (5C2 * 2)/5! = 20/120 = 1/6

Posted from my mobile device

Originally posted by jhavyom on 08 Dec 2019, 05:46.
Last edited by jhavyom on 08 Dec 2019, 22:48, edited 1 time in total.
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Re: Ram had to post 5 letters to 5 different persons. He put the l  [#permalink]

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New post 08 Dec 2019, 06:02
Ram had to post 5 letters to 5 different persons. he will send exactly two letters to correct address i.e out of 5 only 2 will be sent , so the possible outcomes are 5c2= 10
total possible outcome for 5 letter for 5 address would be 5!= 120
probability of sending exactly 2 letters to correct address = 10/120= 1/12
hence answer is A.
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Re: Ram had to post 5 letters to 5 different persons. He put the l  [#permalink]

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New post 08 Dec 2019, 18:45
Ram had to post 5 letters to 5 different persons. He put the letters in the address stamped envelop randomly. What is the probability that exactly two letters are sent to correct address?

The total possibilities = 5!= 120 ways

Exactly two letters are sent to correct address: C_25= 5!/(2!*3!)= 5*4/2=10 ways
--> But, if two letters are sent correctly, other letters should be sent incorrectly.
Let's say that
\(L_1\)-->\(E_1\)
\(L_2\)-->\(E_2\)
---------- CORRECTLY SENT

\(L_3\)--> \(E_4\)
\(L_4\)--> \(E_5\)
\(L_5\)--> \(E_3\)

or
\(L_3\)--> \(E_5\)
\(L_4\)--> \(E_3\)
\(L_5\)--> \(E_4\)
-----------INCORRECTLY SENT

Other 3 letters can be sent incorrectly in 2 ways
--> 10*2=20 ways
Total possibilities which exactly two letters are sent correctly are 20 ways

--> P=\(\frac{20}{120} = \frac{1}{6}\)

The answer is B.
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Re: Ram had to post 5 letters to 5 different persons. He put the l  [#permalink]

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New post 09 Dec 2019, 05:56
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For exactly two letters to be correctly posted, the other three letters must be put in the incorrect envelopes.

Select two letters out of 5 ⇒ 5C2 ways
Put two selected letters in correct envelopes ⇒ 1 way

Put remaining three letters in wrong envelopes ⇒ 2 ways

Total (favourable) ways = 5C2 * 1 * 2 = 20

5 letters can be put in 5 envelopes in 5! ways i. e. 120 ways

P = 20/120 = 1/6
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Re: Ram had to post 5 letters to 5 different persons. He put the l   [#permalink] 09 Dec 2019, 05:56
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