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01 May 2023, 08:00
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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:
67% (02:05) correct
33%
(02:13)
wrong
based on 120
sessions
History
Date
Time
Result
Not Attempted Yet
Find the number of numbers, greater than a million, that can be formed with the digits 2,3,0,3,4,2,3.
a) 240
b) 360
c) 720
d) 480
e) 120
a) 240
b) 360
c) 720
d) 480
e) 120
01 May 2023, 19:17
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Asked: Find the number of numbers, greater than a million, that can be formed with the digits 2,3,0,3,4,2,3.
Number of choices for millionth digit = 6
Number of choices for 100 thousandth digit = 6
Number of choices for 10 thousandth digit = 5
Number of choices for thousandth digit = 4
Number of choices for hundredth digit = 3
Number of choices for tenth digit = 2
Number of choices for unit digit = 1
Since there are two 2s and three 3s, number of numbers must be divided by 2!3!
Number of numbers, greater than a million, that can be formed with the digits 2,3,0,3,4,2,3 = 6*6*5*4*3*2*1/2!3! = 360
IMO B
Number of choices for millionth digit = 6
Number of choices for 100 thousandth digit = 6
Number of choices for 10 thousandth digit = 5
Number of choices for thousandth digit = 4
Number of choices for hundredth digit = 3
Number of choices for tenth digit = 2
Number of choices for unit digit = 1
Since there are two 2s and three 3s, number of numbers must be divided by 2!3!
Number of numbers, greater than a million, that can be formed with the digits 2,3,0,3,4,2,3 = 6*6*5*4*3*2*1/2!3! = 360
IMO B
01 May 2023, 23:23
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CapnSal
More than 1M, 1 000 000, So First digit cant be 0 in the set. All other's are > 1M
So ~Case = Total - Case.
So Total = 7!/3!.2! = 420.
Case where 0 is first, 6!/3!.2! = 60
Hence 360 is desired condition.
