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The sum of the digits in the unit place of all the 4 digit numbers formed with the help of 3, 4, 5 and 6 taken all at a time is:
- a)18
- b)108
- c)432
- d)144
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
The sum of the digits in the unit place of all the 4 digit numbers for...
Required sum = 3!(3 + 4 + 5 + 6) = 108
[If we fix 3 of the unit place, other three digits can be arranged in ways similarly for 4, 5, 6.]
[If we fix 3 of the unit place, other three digits can be arranged in ways similarly for 4, 5, 6.]
Most Upvoted Answer
The sum of the digits in the unit place of all the 4 digit numbers for...
Required sum = 3!(3 + 4 + 5 + 6) = 108
[If we fix 3 of the unit place, other three digits can be arranged in ways similarly for 4, 5, 6.]
[If we fix 3 of the unit place, other three digits can be arranged in ways similarly for 4, 5, 6.]
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Community Answer
The sum of the digits in the unit place of all the 4 digit numbers for...
Using 3,4,5,6 you can make 4 digit words in 4 x 3 x 2 x 1 = 24 ways. So there are 24 numbers in total. If you fix the unit place, you can choose the rest of the places in 3 x 2 x 1 = 6 ways.
So there are 6 numbers each ending with 3,4,5,6.
Sum = ( 6 + 5 + 4 + 3 ) x 6 = 108
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The sum of the digits in the unit place of all the 4 digit numbers formed with the help of 3, 4, 5 and 6 taken all at a time is:a)18b)108c)432d)144Correct answer is option 'B'. Can you explain this answer?
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