How many number less than 10000 and divisible by 5 can be formed with the 10 digit 1,2,3,4,5,6,7,8,9,0 each digit not occurring more than one in each number?
Answers
Answer:
1106 numbers
Step-by-step explanation:
We need the numbers less than 10000, so the numbers until 9999.
The numbers might be 1-digit, 2-digit, 3-digit or 4-digit.
(i) 1-digit numbers divisible by 5:
5(only one number) = 1
(ii) 2-digit numbers divisible by 5:
The unit digit could be 0 or 5 = 2 ways
If the unit digit is 0:
The unit place can be filled in 1 way (0).
The tenths place can be filled in 9P1 ways
So, the number of possible ways = 9P1 x 1 = 9 ways
If the units digit is 5:
The unit place can be filled in 1 way (5).
The tenths place can be filled in 8P1 ways.
So, the number of possible ways = 8P1 x 1 = 8 ways
So, the number of 2-digit numbers possible = 9 + 8 = 17
(iii) 3-digit numbers divisible by 5:
The unit digit could be 0 or 5 = 2 ways
If the units digit is 0:
The unit place can be filled in 1 way (0).
The remaining 2 digits can be filled in 9P2 ways
So, the number of possible ways = 9P2 x 1 = 9 x 8 ways
= 72
If the units digit is 5:
The unit place can be filled in 1 way (5).
The 100's digit can be filled using 1 though 9(except 5)= 8 ways
The remaining 1 digits can be filled in 8 ways
So, the number of possible ways = 8 x 8 x 1 = 64
So, the number of 3-digit numbers possible = 72 + 64
= 136
(iv) 4-digit numbers divisible by 5:
The units digit could be 0 or 5 = 2 ways
If the unit digit is 0:
The unit place can be filled in 1 way (0).
The remaining 3 digits can be filled in 9P3 ways
So, the number of possible ways = 9P3 x 1 = 9 x 8 x 7ways
= 504
If the units digit is 5:
The unit place can be filled in 1 way (5).
The 1000's digit can be filled using 1 though 9(except 5)= 8 ways
The remaining 2 digits can be filled in 8P2 ways
So, the number of possible ways = 8P3 x 1 = 8P2 x 8 x 1 ways
= 8 x 7 x 8 x 1
= 448
So, the number of 4-digit numbers possible = 504 + 448
= 952
The total number of digits under 10000 and divisible by 5 = 1-digit numbers divisible by 5 + 2-digit numbers divisible by 5 + 3-digit numbers divisible by 5 + 4-digit numbers divisible by 5
= 1 + 17 + 136 + 952
= 1106 numbers
Step-by-step explanation:
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