Suppose we uniformly and randomly select permutations from the 20! Permutations of 1, 2, 3,..., 20. What is the probability that 2 appears at an earlier position than any other even number in the selected permutation?
A. 0.5
B. 0.1
C. 9! / 20!
D. None
According to the work book the answer is (D)
However am getting the answer as 0.1
Here goes the explanation.
The first even number is 2. Selection of these numbers is analogies to arranging them in 20 blanks.
- Suppose 2 is filled in the first blank. then the remaining 19 elements can be arranged in 19! Ways.
- Suppose 2 goes into the second blank. The 1st blank can be occupied by an odd number in 10P1 ways and the 2nd blank can be filled with 2 in 1 way while the remaining 18 blanks can be filled in 18! Ways Hence (10P1)(1)(18!) ways
- Suppose 2 goes into the third blank. The 1st two blanks can be occupied by two odd number in 10P2 ways and the 3nd blank can be filled with 2 in 1 way while the remaining 18 blanks can be filled in 17! Ways Hence (10P2)(1)(17!) ways
Generalizing the above three steps.
For i = 0 to 10
Summation of
( 10Pi )(1)(19 - i )!
Evaluating the above summation in the calc, I got 1/10 = 0.1
Let me know if I missed anything in the calculation.
I agree with you, and think there is a much easier explanation. As you fill in the list, if you pick an odd number it doesn't matter. The only question is whether the first even you pick is $2$ or something else. There are $10$ evens to pick, so the chance the first one is $2$ is $1/10$