Find the probability of getting an 8 digit number divisible by 2 using the digits of 1, 2, 3, 5, 7, 0, 4, 6. (digits should not repeat)
Answers
Given : 8 digit number using the digits of 1, 2, 3, 5, 7, 0, 4, 6.
To Find : probability of getting an 8 digit number divisible by 2
Solution:
8 digit number using the digits of 1, 2, 3, 5, 7, 0, 4, 6.
1st Digit can be in 7 ways as 0 can not be 1st digit
remaining 7 digits can be arranged in 7! ways
Hence
7.7! = Total 8 Digit numbers using the given digits
1st Digit is 2 , 4 or 6 in 3 ways
Then last Digit can be in 3 ways ( as 0 , 2 , 4 or 6 only possible for even numbers) and one of the Digit is already used.
remaining 6 digits in between can be in 6 ! ways
3 * 6! * 3
1st Digit is 1 , 3 , 5 , 7 in 4 ways
Last Digit can be in 4 ways ( 0 , 2 , 4 or 6)
remaining 6 digits in between can be in 6 ! ways
4*6!*4
Hence Total 8 digit number divisible by 2 using the give digits
3 * 6! * 3 + 4*6!*4
= 25 * 6!
Probability = 25 * 6! / 7.7!
= 25/49
25/49 is the probability of getting an 8 digit number divisible by 2 using the digits of 1, 2, 3, 5, 7, 0, 4, 6.
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Answer:
5-digit numbers divisible by 5 must end in 0 or 5.
Assuming all numbers to be formed will be 5-digits numbers:
All 5-digit numbers which can be formed from the set [0, 1, 2, 4, 5, 6, 8, 9] without repetition:
7 *7 * 6 * 5 *4 = 5880
All 5-digit numbers divisible by 5 which can be formed from the set [0, 1, 2, 4, 5, 6, 8, 9] without repetition:
Case 1:
5-digits ending in 0 :
7 * 6 * 5 * 4 * 1 = 840
Case 2:
5-digits ending in 5:
6 * 6 * 5 * 4 * 1 = 720
Total number of 5-digit numbers divisible by 5:
840 + 720
= 1560.
Therefore, probability of forming a 5-digit number divisible by 5 :
1560/5880 = 156/588
= 0.265