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26 Mar 2019, 05:38
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
48% (03:09) correct
52%
(02:40)
wrong
based on 114
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How many positive even three-digit integers have the property that their digits, read left to right, are in strictly increasing order?
(A) 21
(B) 34
(C) 54
(D) 72
(E) 150
(A) 21
(B) 34
(C) 54
(D) 72
(E) 150
26 Mar 2019, 05:58
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Bunuel
I am assuming "strictly increasing order" means the digits cannot be equal.
For the numbers to be even, the units digit must be 0/2/4/6/8. But it must be the greatest digit too. So it cannot be 0 or 2.
Units digit 4:
The two leftmost digits can be 1/2/3. Two can be chosen in 3C2 ways and arranged in only 1 way (smaller on left and greater in right)
So 3 ways
Units digit 6:
The two leftmost digits can be 1/2/3/4/5. Two can be chosen in 5C2 ways and arranged in only 1 way (smaller on left and greater in right)
So 5*4/2 = 10 ways
Units digit 8:
The two leftmost digits can be 1/2/3/4/5/6/7. Two can be chosen in 7C2 ways and arranged in only 1 way (smaller on left and greater in right)
So 7*6/2 = 21 ways
Total = 3 + 10 + 21 = 34 ways
Answer (B)
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26 Mar 2019, 06:13
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Assume that the first digit is 1 and second digit is 2 then we have 3 possible third digit. While first digit remains 1 second changes to 3 then we have 3 possible third digit. So with first digit 1 we have 3,3, 2,2,1,1,possible numbers when we change first digit to 2 we will have 3,2,2,1,1,possible numbers and so on. So 12+9+6+4+2+1=34 option B.
Is there a shorter way to work it out?
Posted from my mobile device
Is there a shorter way to work it out?
Posted from my mobile device
