♦ For the first question, "How many numbers of 3 digits can be formed with 0,2,4,6,8?"
Since for it to be a 3 digit number the hundreds place shouldn't be 0.
Now, the no. of digits that can be placed in the ones place is 5, since there are 5 digits.
Similarly for the tens place, it is 5 again. But for the hunderds place it is 4, as 0 is among these 5 digits and it cannot be counted.
So, the no. of 3 digit numbers possible is ⇒ 4x5x5 = 100.
♦ Now, the second question,
"Numbers are formed by using all the digits 1,2,5,6,9.
Each digit appearing only once in each number, how many of these will be:
(a) divisible by 5
(b) divisible by 25
(c) odd
(d) even
(e) greater than 50,000"
Now lets find how many numbers can be formed,
Here, 1,2,5,6,9 are given to us, and we are supposed to use all the digits.
We can use the same logic as the first question, but here the repetition of a single digit is not allowed, as are supposed to use all the digits.
Now we can use Permutations, since there are 5 digits, the no. of numbers formed from these digits is 5! = 120.
(a) Now, for the number to be divisible by 5, the number should end with either 0 or 5.
Since there is no 0 and only 5 is present. We lock the last digit to be 5 and use Permutation.
_ _ _ _ 5, we can represent it like this. Now the no. ways we arrange the remaining numbers in the 4 blank spaces is ⇒ 4! = 24
The no. of numbers divisible by 5 = 24
(b) Now for a number to be divisible by 25, the last 2 digits of the number should be 00,25,50,75.
since there is only 2 & 5 present, the last 2 digits should be 25.
_ _ _ 25, we can represent it like this, now the no. ways we arrange the remaining numbers in the 3 blank spaces is ⇒ 3! = 6
The no. of numbers divisible by 25 = 6
(c) & (d)
For a number to be odd, it should end with an odd number obviously.
So there are 3 odd numbers.
They can be represented as ⇒ _ _ _ _ 1, _ _ _ _ 5 & _ _ _ _ 9
Using the permutation logic like we used in (a) & (b).
The no. ways we arrange the remaining numbers in the 4 blank spaces of each ⇒ 4!
So the total no. of odd number is ⇒ 4! + 4! + 4! = 24+24+24 = 72.
Now, if 72 numbers are odd, the remaining numbers left are even obviously.
The total no. of even numbers = 120 - 72 = 48.
(Total numbers are 120 as we found earlier.)
(e) For a number to be greater than 50,000 among these 120 numbers, the first digit of the number should be either 5,6 or 9. Similarly like we did for (c),
we can represent them as ⇒ 5 _ _ _ _ , 6 _ _ _ _, & 9 _ _ _ _.
Using the permutation logic, The no. ways we arrange the remaining numbers in the 4 blank spaces of each ⇒ 4!
So, the total no. of numbers greater than 50,000 is ⇒ 4! + 4! + 4! = 24+24+24 = 72.
