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17 Nov 2020, 08:00
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
71% (02:00) correct
29%
(02:28)
wrong
based on 120
sessions
History
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Time
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Not Attempted Yet
How many positive integers consisting of five different digits can be made from digits 1, 2, 3, 4, 5, 6, 7, 8, and 9 if the numbers must be odd ?
A. 2520
B. 5040
C. 8400
D. 13440
E. 15120
A. 2520
B. 5040
C. 8400
D. 13440
E. 15120
17 Nov 2020, 13:55
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Bunuel
We first pick an odd number for the units digit. We can choose from 1, 3, 5, 7, 9 so there are five options for the units digit.
Next we can pick any number we'd like for the rest of the digits. The numbers can't repeat so there are 8 options (one less option as we needed to pick the units digit first) for the ten-thousandths digit, 7 options for the thousandths digit etc. Thus in result we have 8*7*6*5 for the first four digits after we pick one of the 5 possible units digits. \(5*8*7*6*5 = 40*7*30 = 1200*7=8400\).
Ans: C
17 Nov 2020, 15:44
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I used the SLOT method here.
We are given one condition in that the number is odd. This means that the last slot has 5 choices: 1,3,5,7,9
Since we have a condition, we fill the slot with the condition first.
WIth the last number filled, we start with 8 choices for the rest of the slots.
The total is 8 x 7 x 6 x 5 x 5 = 8400
Answer: C
We are given one condition in that the number is odd. This means that the last slot has 5 choices: 1,3,5,7,9
Since we have a condition, we fill the slot with the condition first.
WIth the last number filled, we start with 8 choices for the rest of the slots.
The total is 8 x 7 x 6 x 5 x 5 = 8400
Answer: C
