The Wrong Method
Let
a,
b,
c,
d be the number of experts who get the versions A, B , C and D respectively. Then,
a + b + c+ d = 6 (1)
If we take non-negative integer solutions of (1), we get all the possible cases, which is 9C3
If we take only positive integer solutions of (1), we get all the cases in which all the versions will be available with Alpha, which is 5C3
Hence, the required probability is 9C3/5C3 = 5/42
Hence, option 2.
The Correct Method
Total number of ways of getting the versions, 6 people can each get any of the four versions = 46
Then, call the experts P, Q, R, S, T, U.
Then, for all the versions to be present, each of them are present and the rest two can be either same or different.
When the fifth and sixth are same, then we this can be done in 4(6!/3!) = 480 ways.
When the fifth and sixth are different, the two which are repeated can be selected in 4C2 = 6 ways.
Hence, total number of ways = 6(6!/(2!2!)) = 1080
Hence, the required probability = (1080 + 480)/4096 = 195/512
Hence, option 4.
