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My high school geometry textbook includes the following problem as an example:

Three pool balls are chosen at random from a set numbered from 1 to 15. What is the probability that the pool balls chosen are numbered 5, 7, and 9?

My answer is $\frac{1}{455}$. The text book's answer is $\frac{6}{455}$. Which answer is correct?

I arrive at my answer, as follows:

  1. The probability of the desired result is the ratio of the number of outcomes that yield the desired result to the total number of possible outcomes.

  2. The desired result, balls 5, 7 and 9, together form a single 3-ball combination; a single outcome: $$ 1 $$

  3. The total number of possible 3-ball combinations is: $$ \frac{15!}{3!(15-3)!} = \frac{(15 \cdot 14 \cdot 13)}{6} = \frac{2730}{6} = 455 $$

  4. Thus, the ratio described in (1), above, is $1\colon455$, and the answer is: $$ \frac{1}{455} $$

There are at least two alternative approaches that yield the same answer:

  • Rather than using combinations, one might use permutations, in which case the numerator is the $3!$ permutations that yield the desired result, and the denominator is the $\frac{15!}{(15-3)!}$ permutations that could occur in choosing 3 balls, which is equal to $\frac{6}{2730}$ = $\frac{1}{455}$.
  • Or, one could use the product of the odds of each selection being a success: $\frac{3}{15} \cdot \frac{2}{14} \cdot \frac{1}{13}$ = $\frac{6}{2730}$ = $\frac{1}{455}$.

HOWEVER, my textbook says the solution is: $$ \frac{3!}{\frac{15!}{3!(15 - 3)!}} = \frac{6}{455} $$

The textbook's statement of the problem and solution is here: Lesson 13-3, Permutations and Combinations: Problem 6. I believe the textbook's solution compares a number of permutations in the numerator to a number of combinations in the denominator, which seems to me to be a matter of apples and oranges.

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  • $\begingroup$ Welcome to MathSE. You are correct. Please notify the publisher about the error. $\endgroup$ Commented Sep 11, 2018 at 0:34
  • $\begingroup$ Yes, one possibility works among the $\binom {15}3=455$, the probability is $1/455$. $\endgroup$
    – dan_fulea
    Commented Sep 11, 2018 at 0:49
  • $\begingroup$ @N.F.Taussig, thank you for the kind welcome. This looks like the right place for someone who enjoys math. $\endgroup$ Commented Sep 11, 2018 at 1:19

1 Answer 1

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Your answer is correct. We have to consider either permutation or combination for both numerator and denominator to get the correct probability.

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  • $\begingroup$ I appreciate you taking the time to provide an answer. Thanks. $\endgroup$ Commented Sep 11, 2018 at 1:17

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