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How many numbers with 3 different digits can...

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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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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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