If 5 integers are to be selected at random and without replacement

 
Probability approach:

We need the probability of {10, 20, any, any, any}.

P(10, 20, any, any, any) = 1/20 * 1/19 * 1 * 1 * 1 * 5!/3! = 1/19. We multiply by 5!/3! to account for all permutations of {10, 20, any, any, any} (such as {10, 20, any, any, any}, {10, any, 20, any, any}, {10, any, any, 20, any}, ...).

Answer: D.­

To elaborate more:

• Case 1: {10, 20, any, any, any}. P = 1/20 * 1/19 * 1 * 1 * 1
• Case 2: {10, any, 20, any, any}. P = 1/20 * 18/19 * 1/18 * 1 * 1
• Case 3: {10, any, any, 20, any}. P = 1/20 * 18/19 * 17/18 * 1/17 * 1
• ...
• Case 20: {any, any, any, 20, 10}. P = 18/20 * 17/19 * 16/18 * 1/17 * 1/16

Here, when we say "any," it denotes "any number other than 10 or 20". It's important to note that the probability for each case is essentially 1/20 * 1/19. Given that there are a total of 20 cases, the overall probability becomes 1/20 * 1/19 * 20 = 1/19.

Combination approach:

The total number of ways to select 5 different numbers from 20 is 20C5.

The number of ways to select 10 and 20 is 1, and the total number of ways to select the remaining 3 numbers from 18 is 18C3.

Hence, the probability is:

\(\frac{18C3}{20C5} = \frac{(\frac{18!}{15!3!})}{(\frac{20!}{15!5!})} = \frac{18!}{15!3!}*\frac{15!5!}{20!}=\frac{1}{19}\)

­Answer: D.­