How many 3-letter codes can be formed by choosing, without replacement, 3 letters from the word PEPPER?

Correct answer marked below

A

6

B

18

19

Correct

D

27

E

30

You can see the correct answer above. Sign in for free to unlock the complete worked solution.

Get detailed analytics, unlock full explanations, and move up difficulty tiers as you practice.

Explanation

Preview

Official Explanation

Official

Only these cases are possible:

PER

= 3! = 6

EER

= \frac{3!}{2!} = 3

EEP

= \frac{3!}{2!} = 3

PPE

= \frac{3!}{2!} = 3

…

Sign in for free to see the complete step-by-step solution to this question.

Comments

Sign in to join the discussion

Be the first to share your thoughts and help others understand it better.

What this question tests

Whether you can count arrangements and selections correctly — permutations vs. combinations.

Overcounting: treating a selection as ordered when order does not matter, or forgetting to divide by duplicates.

Create a free account to reveal the correct answer and see step-by-step explanations for every question.