I have another problem solving question....?

Question

How many 3-digit numbers can be formed from the digits 0, 1, 2, 3, 4, 5? How many of them are odd? How many are even?

Answer 1

Since the question asks about 3-digit *numbers* (rather than a sequence of digits), I assume that you can't have a leading zero.

First digit can be 1-5 (5 choices)
Second digit can be 0-5 (6 choices)
Third digit can be 0-5 (6 choices)

5 x 6 x 6 = 180 combinations

As for how many are odd or even, they are split evenly. Half end with 0,2,4 and are even. Half end with 1,3,5 and are odd. (90 odd, 90 even).

Note: if you aren't allowed to repeat digits, then this answer would be different, but the question doesn't seem to disallow 3-digit numbers like 100, 522, 333, etc.

Answer 2

The first digit of the three digit number can be any of the six numbers 0,1,2,3,4,5; whichever choice is made, the second digit can also be any of those six numbers, giving 6x6=36 combinations of the first two digits; and whichever choices are made for the first two digits, the third digit can also be any of the same six numbers, so the total number of three digit numbers that can be made is 36x6=216. Those with the third digit as 0, 2 or 4 number 36x3=108 and are even, while those that end in 1, 3 or 5 make up the other 108 and are odd.

PS In the above solution I have allowed numbers like 001 and 012 to be counted as three digit numbers, as would be the case in most practical situations such as telephone numbers, security codes or lock combinations. However, if leading zeroes are not permitted, then there are only five choices for the first digit, 1,2,3,4,5, and so 5x6x6=180 numbers altogether, again of which half are even and half odd. Speak to your teacher about the ambiguity of the term "3-digit number".

Answer 3

We have 6 digits 0, 1, 2, 3, 4, 5

we can make 6P3, 3 digit numbers i.e. 6!/3! = 6x5x4 = 120 numbers.

for a number to be even the last digit must get divided by 2 so we are left with digits 0, 2 & 4

now we can make 6P3/3P1 = 120/3! = 120/6 = 20 numbers

for a number to be odd the last digit must not get divided by 2 so we are left with digits 1, 3 & 5

now we can make 6P3/3P1 = 120/3! = 120/6 = 20 numbers