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07 May 2020, 10:29
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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:
55%
(hard)
Question Stats:
51% (01:31) correct
49%
(01:42)
wrong
based on 116
sessions
History
Date
Time
Result
Not Attempted Yet
How many integers, greater than 999 but not greater than 4000, can be formed with the digits 0, 1, 2, 3 and 4, if repetition of digits is allowed?
(A) 375
(B) 376
(C) 499
(D) 500
(E) 501
(A) 375
(B) 376
(C) 499
(D) 500
(E) 501
Updated on: 07 May 2020, 20:38
Originally posted by GMATinsight on 07 May 2020, 10:40.
Last edited by GMATinsight on 07 May 2020, 20:38, edited 1 time in total.
Last edited by GMATinsight on 07 May 2020, 20:38, edited 1 time in total.
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Bunuel
999 but not greater than 4000 i.e. all 4 digit numbers less than 4000
Digits to be used are 0, 1, 2, 3 and 4 (if repetition of digits is allowed)
We have to fill these 4 empty spaced for digits tomake number
_ _ _ _
The left most place can be filled in 3 ways using digits 3, 2 and 1 only
Second left most place can be filled in 5 ways using digits 0, 4, 3, 1 and 2
Tens digit place can be filled in 5 ways using digits 0, 4, 3, 1 and 2
Unit digit place can be filled in 5 ways using digits 0, 4, 3, 1 and 2
Total Numbers = 3*5*5*5 = 375
But Since 4000 is included so an extra count = 1
Total Favorable cases = 375+1 = 376
Answer: Option B
07 May 2020, 11:22
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Answer: B
We need integers greater than 999 but not greater than 4000. So from 1000 to 4000 (inclusive).
Digits available: 0,1,2,3,4
_ _ _ _
Leftmost digit can have values 1,2,3 = 3 ways
2nd digit from left can take all 5 digit as values = 5 ways
3rd digit from left can take all 5 digit as values = 5 ways
rightmost digit can take all 5 digits as values = 5 ways
Total = 3*5*5*5=375
Since its mentioned integers not greater than 4000, we need to consider 4000 as well.In the total above, we haven't counted 4000 (we haven't considered digit 4 in the leftmost position).
So 1 more integer will be counted for the value 4000,
Total integers = 375 +1 = 376
We need integers greater than 999 but not greater than 4000. So from 1000 to 4000 (inclusive).
Digits available: 0,1,2,3,4
_ _ _ _
Leftmost digit can have values 1,2,3 = 3 ways
2nd digit from left can take all 5 digit as values = 5 ways
3rd digit from left can take all 5 digit as values = 5 ways
rightmost digit can take all 5 digits as values = 5 ways
Total = 3*5*5*5=375
Since its mentioned integers not greater than 4000, we need to consider 4000 as well.In the total above, we haven't counted 4000 (we haven't considered digit 4 in the leftmost position).
So 1 more integer will be counted for the value 4000,
Total integers = 375 +1 = 376
