(a) How many three-digit numbers can be formed from the digits 0, 1, 2, 3, 4, 5, and 6 if
each digit can be used only once?
(b) How many of these are odd numbers?
(c) How many are greater than 330?
This is what I have done:
a) We can choose from 6 numbers for the first digit ( we exclude 0), 6 digits for the second (we exclude the first but include 0) and finally 5 digits for the third (we exclude the first and second). So total number of$$ \text{possibilities} = 6 \cdot 6 \cdot 5 = 180 \text{ways}$$
b) I have no idea how to approach this. How can we do this?
c) I considered the case when the first digit is 3 , then for the second digit we have the possibilities of {4,5,6} and the last digit {0,1,3,4,5,6}. However we exclude 3, and one more number that has been chosen as the 2nd digit for our last number. So the number of $$\text{possibilities} = 1\cdot 3 \cdot (7-2) = 15$$
Now I considered when the first digit is greater than 3, {4,5,6} then for the second digit we can use {0,1,2,3,4,5,6} (but we exclude the number that has been used as the first digit). Finally for the third {0,1,2,3,4,5,6} and we exclude 2 numbers than have been used. So the number of $$ \text{possibilities} = 3 \cdot (7-1) \cdot (7-2) =90$$
In conclusion we have: $$90 + 15 = 105$$ total possibilities greater than 330.
Thank you for your time!
Best Answer
a) $6\cdot 6\cdot 5$ is correctly the way to select three digit numbers from that set of seven digits when excluding the possibility of leading zeros. $180$
b) There are $3$ odd digits for the units, now how many 2 digit numbers can be made from the remaining 6 digits, excluding leading zeros? (The second verse is same as the first!)
c) To be greater than 330 with those digits we do case work. Either we start with a 3, or we start higher. As you did. $1\cdot 3\cdot 5+ 3\cdot 6 \cdot 5 = 105$