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.
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How many numbers with 3 different digits can...
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BTGmoderatorLU
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The first place of the three digit number can be filled with any of the five numbers 1,3,5,7,9LUANDATO 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.
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
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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: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.
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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