The first boy enters the room and found 6 hats. He gets

Question

The first boy enters the room and found 6 hats. He takes one and then the second boy enters and also gets one hat. And third boy come and also gets one hat. The next day, three of them find out that they got someone else’s hat. In how many ways they could have gotten the wrong hats?

A. 120
B. 75
C. 71
D. 68
E. 60

A. 120
B. 75
C. 71
D. 68
E. 60

nerdynerdnerd · Answer

When boy 1 comes into the room, 5 out of 6 hats are not his
for boy 2, 4 out of 5 hats are not his
for boy 3, 3 out of 4 hats are not his.

For all 3 of them to get the wrong hats = 5 x 4 x 3 = 60

shiva1683 · Answer

What about the situations when the boy 1 took the boy 2’s / boy 3’s hat and similarly the boy 2 took boy 3s hat? That would imply they can go wrong in more ways, right?

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nick1816 · Answer

N(a), Number of cases when first boy has his cap= 1*5*4=20 
N(b), Number of cases when second boy has his cap= 5*1*4=20 
N(c), Number of cases when third boy has his cap= 5*4*1=20

N(A∩B), Number of cases when first and second boy have their cap= 1*1*4=4
N(B∩C), Number of cases when second and third boy have their cap= 4*1*1=4
N(C∩A), Number of cases when first and third boy have their cap= 1*4*1=4

N(A∩B∩C), Number of cases when first, second and third boy have their cap= 1*1*1=1

Total number of cases in which atleast 1 person has its cap= 20+20+20-4-4-4+1=49

Total possible cases= 6C3*3!=120

Total number of cases in which no one has its cap= 120-49=71

chetan2u  I think OA is not correct. What do you think?

chetan2u · Answer

Yes  nick1816  The OA is wrong and, also, the question is poorly worded.

We can do it in two ways..

(I) Finding scenarios where any of three get one's hat.. 
(a) all three get their hat ---1 way
(b) Two of them get their hat --
Say A and B get their hat while C doesn't, so C can choose from remaining 4 minus his own -- 3 ways
Similarly for A and B not getting , 3 each. Total 3*3=9 ways
(c) Only one getting his hat
Say A gets his hat
*B gets C's hat and then C can get remaining 4 -- 4 ways
*B gets other than C's hat, so 3 ways, and then C can get remaining 4 minus his own -- 3 ways...Total 3*3=9 ways
Total for each -4+9=13 ways
similarly 13 each for other two, that is when only B gets his hat and only when only C gets his hat, so 13*3=39 ways

Total - 1+9+39=49 ways..Our answer = total-49=6*5*4-49=71ways

(II) Direct method 
(a) All 3 getting each others hat - Only 1 way
(b) Two of their hats between each other, --Total 3 ways
(c) Only 1 of their hat amongst each other

hr1212 · Answer

Let the boys be A, B and C then total ways they can select hats = 6C1*5C1*4C1 = 120

Now by the law of probability, no. of ways in which A can get his own hat =  \frac{120}{6}  = 20 ways

Number of ways each of these three get their own hats P(A and B and C) = 1
Number of ways 2 of them get their own hats = select 2 boys and give them their own hats (3C2) * select 1 hat for the third boy which is not his own hat (6 - 2 - 1 = 3) = 3 * 3 = 9

Again by the law of probability, number of ways (A and B) or (B and C) or (A and C) get their own hats = Total/possibilities =  \frac{9}{3 }  = 3 ways (Calculated just to visualize the Venn diagram).

Number of ways out of 20, where A gets his own hat P(only A) = P(A) - P(A and B) - P(A and C) - P(A and B and C) = 20 - 3 - 3 - 1 = 13

We can also plot the Venn diagram for better understanding, where either one of these boys get their own hats.

Total number of ways from Venn diagram where either of these boys get their own hats = 13 + 13 + 13 + 3 + 3 + 3 + 1 = 49

Total ways where neither of them get their own hat = Total ways - Total ways where either of them get their own hat = 120 - 49 = 71

Answer: C

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