How many numbers with 3 different digits can...

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BTGmoderatorLU: Moderator; Posts: 2650; Joined: Sun Oct 15, 2017 1:50 pmFollowed by: 6 members

by BTGmoderatorLU » Fri Jan 12, 2018 2:06 pm

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How many numbers with 3 different digits can be formed using only the odd digits?

(A) 10
(B) 20
(C) 40
(D) 60
(E) 120

The OA is D.

I'm confused with this PS question. Experts, any suggestion? How can I solve this question? Thanks in advance.

Source: Beat The GMAT — Problem Solving | Fri Jan 12, 2018 2:06 pm

GMATWisdom: Master | Next Rank: 500 Posts; Posts: 100; Joined: Wed Nov 29, 2017 4:38 pm; Thanked: 14 times

by GMATWisdom » Fri Jan 12, 2018 2:22 pm

LUANDATO wrote:How many numbers with 3 different digits can be formed using only the odd digits?

(A) 10
(B) 20
(C) 40
(D) 60
(E) 120

The OA is D.

I'm confused with this PS question. Experts, any suggestion? How can I solve this question? Thanks in advance.

The first place of the three digit number can be filled with any of the five numbers 1,3,5,7,9
second place can be filled by any of the remaining four numbers and
the last place could be filled with any of the remaining three numbers.
thus the total numbers formed without repetition of the digits will be 5x4x3=60
option D

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8246; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Tue Aug 06, 2019 4:34 pm

BTGmoderatorLU wrote:How many numbers with 3 different digits can be formed using only the odd digits?

(A) 10
(B) 20
(C) 40
(D) 60
(E) 120

The OA is D.

The first digit has 5 choices (1, 3, 5, 7, 9), then the second and third digits have 4 and 3 choices, respectively. So the total number of numbers with 3 different odd digits is:

5 x 4 x 3 = 60

Alternate Solution:

The three digits can be chosen in 5C3 = 5!/(3!*2!) = (5 x 4)/(2 x 1) = 5 x 2 = 10 ways. The chosen digits can be arranged in 3! = 3 x 2 = 6 ways. Thus, a total of 10 x 6 = 60 different numbers is possible.

Answer: D

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