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01 Feb 2022, 04:00
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
59% (02:25) correct
41%
(02:11)
wrong
based on 102
sessions
History
Date
Time
Result
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How many even numbers of 4 digits can be formed using digits 0, 1, 2, 3, 4, 5 and 6 ? (Repetition is not allowed)
A. 420
B. 480
C. 828
D. 840
E. 864
A. 420
B. 480
C. 828
D. 840
E. 864
01 Feb 2022, 22:37
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We have digits 0, 1, 2, 3, 4, 5, 6
We require a 4-digit even number without repetition. So, the last digit must be an even number.
Since we have 0 as one of the digits, let's split them into 2 cases.
Case-1: Assuming 0 occupied at the end
_ x _ x _ x _
Since, 0 present at the last digit, the last digit can be filled in 1 way.
_ x _ x _ x 1
1st digit can be filled in 6 ways (any one digit from 1-6)
6 x _ x _ x 1
2nd digit can be filled in 5 ways (any one digit from the remaining 5)
6 x 5 x _ x 1
3rd digit can be filled in 4 ways (any one digit from the remaining 4)
6 x 5 x 4 x 1
So we have 6x5x4x1 = 120 possibilities
Case-2: Assuming a non-zero digit at the end
_ x _ x _ x _
Since the number is even, the last digit can be filled in 3 ways (2 or 4 or 6)
1st digit can be filled in 5 ways (any digit except 0 and the digit occupied in the 4th position)
Note: A number cannot start with a 0.
2nd digit can be filled in 5 ways again (0 or any other digit except the digit occupied in the 1st and 4th position)
3rd digit can be filled in 4 ways (any digit except the digit occupied in the 1st, 2nd and 4th position)
So we have 5x5x4x3 = 300 possibilities
So we have 120+300 = 420 possibilities in total.
We require a 4-digit even number without repetition. So, the last digit must be an even number.
Since we have 0 as one of the digits, let's split them into 2 cases.
Case-1: Assuming 0 occupied at the end
_ x _ x _ x _
Since, 0 present at the last digit, the last digit can be filled in 1 way.
_ x _ x _ x 1
1st digit can be filled in 6 ways (any one digit from 1-6)
6 x _ x _ x 1
2nd digit can be filled in 5 ways (any one digit from the remaining 5)
6 x 5 x _ x 1
3rd digit can be filled in 4 ways (any one digit from the remaining 4)
6 x 5 x 4 x 1
So we have 6x5x4x1 = 120 possibilities
Case-2: Assuming a non-zero digit at the end
_ x _ x _ x _
Since the number is even, the last digit can be filled in 3 ways (2 or 4 or 6)
1st digit can be filled in 5 ways (any digit except 0 and the digit occupied in the 4th position)
Note: A number cannot start with a 0.
2nd digit can be filled in 5 ways again (0 or any other digit except the digit occupied in the 1st and 4th position)
3rd digit can be filled in 4 ways (any digit except the digit occupied in the 1st, 2nd and 4th position)
So we have 5x5x4x3 = 300 possibilities
So we have 120+300 = 420 possibilities in total.
01 Feb 2022, 23:14
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Easy question with a trick which you could fall for if you're not careful.
So lets start by seeing what the question is asking of us, how many 4 digit even numbers can we form from the set of 0->6
I.e
_ _ _ _
from:
{0,1,2,3,4,5,6}
Well we know an even number must end in either 0,2,4,6 .
so now we can form the following four sets:
_ _ _ 0 --> 1240 , 1350 , 1560 ....
_ _ _ 2 -->4532, 1562, ..
_ _ _ 4
_ _ _ 6
With 3 vacancies in each, and 6 numbers to choose from , each set has 6*5*4 variations => 120/set . 480 in total .
Is that the solution? not yet..
the trick lies in the statement of '4 digit numbers' , whereas by default whenever the 0 will fall in the first digit, the set of numbers will be 3 digits only. ie. 0352 is a 3 digit even number,
So from the total we got earlier we need to subtract the instances from which we have a 0 at the beginning
_ _ _ 0 --> here we're fine because we locked 0 at the last digit and repetition is not allowed
_ _ _ 2 --> here we could have 0 _ _ 2 , which happens 5*4 times--> remove 20 possibilities
_ _ _ 4 --> here we could have 0 _ _ 4 , which happens 5*4 times--> remove 20 possibilities
_ _ _ 6 --> here we could have 0 _ _ 6 , which happens 5*4 times--> remove 20 possibilities
Hence: 480 [total number of 4 digit even numbers] - 60 [number of 3 digit even numbers] = 420.
Hope the explanation was clear
So lets start by seeing what the question is asking of us, how many 4 digit even numbers can we form from the set of 0->6
I.e
_ _ _ _
from:
{0,1,2,3,4,5,6}
Well we know an even number must end in either 0,2,4,6 .
so now we can form the following four sets:
_ _ _ 0 --> 1240 , 1350 , 1560 ....
_ _ _ 2 -->4532, 1562, ..
_ _ _ 4
_ _ _ 6
With 3 vacancies in each, and 6 numbers to choose from , each set has 6*5*4 variations => 120/set . 480 in total .
Is that the solution? not yet..
the trick lies in the statement of '4 digit numbers' , whereas by default whenever the 0 will fall in the first digit, the set of numbers will be 3 digits only. ie. 0352 is a 3 digit even number,
So from the total we got earlier we need to subtract the instances from which we have a 0 at the beginning
_ _ _ 0 --> here we're fine because we locked 0 at the last digit and repetition is not allowed
_ _ _ 2 --> here we could have 0 _ _ 2 , which happens 5*4 times--> remove 20 possibilities
_ _ _ 4 --> here we could have 0 _ _ 4 , which happens 5*4 times--> remove 20 possibilities
_ _ _ 6 --> here we could have 0 _ _ 6 , which happens 5*4 times--> remove 20 possibilities
Hence: 480 [total number of 4 digit even numbers] - 60 [number of 3 digit even numbers] = 420.
Hope the explanation was clear
