Five digit number divisible by 3 is formed using 0,1,2,3,4,6 and 7 without repetition. Total number of such numbers are?:
(1)312
(2)3125
(3)120
(4)216
My answer is coming out to be 504. I don't know where I am going wrong.
Please help
Answers
ans is 216
first digit. It can be chosen 4 ways, so
choose the first digit 4 ways. (It CANNOT be 0)
Choose the second digit 4 ways. (It CAN be 0)
Choose the third digit 3 ways.
Choose the fourth digit 2 ways.
Choose the fifth digit 1 way.
4×4×3×2×1 = 4×4! = 96
Answer: 120+96 = 216
first digit. It can be chosen 4 ways, so
choose the first digit 4 ways.
by 3, because there is no other way to make the
sum 12 or 15, and we certainly can't have a sum of
9 or 18.
600 Five digit number divisible by 3 can be formed using 0,1,2,3,4,6 and 7 without repetition
Step-by-step explanation:
Five digit number divisible by 3 is formed using 0,1,2,3,4,6 and 7 without repetition.
5 numbers out of 7 can be selected in ⁷C₅ = 21 ways
But out of 21 ways
as sum of all 7bdigits = 0+ 1+ 2 + 3 + 4 + 6 + 7 = 23
We need 5 digits hence we need to leave 2 Digit with sum = 2 , 5 , 8 , 11 to have remaining 5 Digits with sum divisible by 3
0 + 2 = 2 , 1 + 4 = 2 + 3 = 5 , 1+7 = 2 + 6 = 8 , 4 + 7 = 11
Sum is divisible by 3 in below cases only
01236 - Numbers can be formed = 4 * 4! = 96
04236 - Numbers can be formed = 4 * 4! = 96
07236 - Numbers can be formed = 4 * 4! = 96
01347 - Numbers can be formed = 4 * 4! = 96
01467 - Numbers can be formed = 4 * 4! = 96
13467 - Numbers can be formed = 5! = 120
Total Number = 96 + 96 + 96 + 96 + 96 + 120 = 600
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