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22 Sep 2021, 03:15
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
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Question Stats:
40% (02:28) correct
60%
(02:33)
wrong
based on 104
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How many integers between 4000 and 7000 have digits that are all different and that increase from left to right?
A. 10
B. 12
C. 14
D. 15
E. 16
A. 10
B. 12
C. 14
D. 15
E. 16
22 Sep 2021, 09:56
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Bunuel
We can distribute these as per the digit in thousands place.
1) 4000 - 4999
First digit is 4, and remaining three can be chosen from 5 to 9 in 5C3 or 10 ways.
2) 5000 - 5999
First digit is 5, and remaining three can be chosen from 6 to 9 in 4C3 or 4 ways.
3) 6000 - 6999
First digit is 6, and remaining three can be chosen in only one way, 789.
Total 10+4+1 = 15 ways.
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Ashitha006
Joined: 29 Jan 2020
Last visit: 20 Jul 2023
Posts: 17
Own Kudos:
Given Kudos: 34
Location: India
Schools: ISB '24 (A)
GMAT 1: 690 Q48 V36
GPA: 3.33
06 Oct 2021, 20:53
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Hi ,
I am getting confused here.
If we are to take each digit different and increase from left to right ,
Case 1 from 4000 to 4999
_ _ _ _
1000 place -1 possible value(4)
100 place-5 possible values (5,6,7,8,9)
10 place-4 possible values (6,7,8,9)
1 place-3 possible values (7,8,9)
Total-5*4*3=60
Case 2 from 5000 to 5999
_ _ _ _
1000 place -1 possible value(5)
100 place-4 possible values (6,7,8,9)
10 place-3 possible values (7,8,9)
1 place-2 possible values (8,9)
Total-4*3*2-24
Case 3 from 6000 to 6999
_ _ _ _
1000 place -1 possible value(6)
100 place-3 possible values (7,8,9)
10 place-2 possible values (8,9)
1 place-1 possible value(9)
Total-3*2*1=6
Case1+case 2+ case 3=60+24+6=90
Please correct me if I went wrong somewhere
Posted from my mobile device
I am getting confused here.
If we are to take each digit different and increase from left to right ,
Case 1 from 4000 to 4999
_ _ _ _
1000 place -1 possible value(4)
100 place-5 possible values (5,6,7,8,9)
10 place-4 possible values (6,7,8,9)
1 place-3 possible values (7,8,9)
Total-5*4*3=60
Case 2 from 5000 to 5999
_ _ _ _
1000 place -1 possible value(5)
100 place-4 possible values (6,7,8,9)
10 place-3 possible values (7,8,9)
1 place-2 possible values (8,9)
Total-4*3*2-24
Case 3 from 6000 to 6999
_ _ _ _
1000 place -1 possible value(6)
100 place-3 possible values (7,8,9)
10 place-2 possible values (8,9)
1 place-1 possible value(9)
Total-3*2*1=6
Case1+case 2+ case 3=60+24+6=90
Please correct me if I went wrong somewhere
Posted from my mobile device
