twobagels wrote:
An n-sided dice, having numbers from 1 to n, is rolled 3 times. If the number of ways of obtaining the same number in exactly two of the three rolls is between 120 and 130, what could the value of n be?
A. 6
B. 7
C. 8
D. 9
E. 10
Solution:
If n = 6 (i.e., the die is a regular 6-sided die), we know there are 216 outcomes when it is rolled 3 times. Now, let’s see how many of these outcomes have exactly 2 of the 3 numbers the same.
Let’s say the two distinct numbers are 1 and 2, so we can have:
1-1-2, 1-2-1, 2-1-1, 2-2-1, 2-1-2 and 1-2-2
We see that there are 6 ways to have exactly 2 of the 3 numbers the same when the 3 numbers consist only of two distinct numbers 1 and 2. However, since there are 6C2 = (6 x 5) / 2 = 15 ways to choose 2 distinct numbers from 6, there are a total of 6 x 15 = 90 ways to have exactly 2 of the 3 numbers the same. Of course, since we are given that the total number of ways should be between 120 and 130, we see that n can’t be 6.
Now, let’s say n = 7. For every pair of two distinct numbers chosen, there are still 6 ways to have exactly 2 of the 3 numbers the same. Since there are 7C2 = (7 x 6) / 2 = 21 ways to choose 2 distinct numbers from 7, there are a total of 6 x 21 = 126 ways to have exactly 2 of the 3 numbers the same. We see that 126 is between 120 and 130, so n must be 7.
Answer: B _________________
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