Without repeating any digit, how many 3-digit numbers divisible by 5 can be formed from the digits 2, 3, 5, 6, 7, and 9?
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☆ranshsangwan☆
As 3 digit number is divisible by 5, its unit place should contain digits 0 or 5. As you can see we have only 5 present here, put it in units place. Now you have 5 digits - 2, 3, 6, 7, 9 remaining.
So you can select 5 digits for tens place and as repeatation is not allowed, for hundreds place you can select remaining 4 digits. So final answer is 4 * 5 * 1 = 20 numbers can be formed.
OR
You can select 2 digits out of 5 digits given in C(5,2) = 20 ways. And as units place contain only 5, final answer becomes 20 * 1 = 20.
As 3 digit number is divisible by 5, its unit place should contain digits 0 or 5. As you can see we have only 5 present here, put it in units place. Now you have 5 digits - 2, 3, 6, 7, 9 remaining.
So you can select 5 digits for tens place and as repeatation is not allowed, for hundreds place you can select remaining 4 digits. So final answer is 4 * 5 * 1 = 20 numbers can be formed.
OR
You can select 2 digits out of 5 digits given in C(5,2) = 20 ways. And as units place contain only 5, final answer becomes 20 * 1 = 20.
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