(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!

  • $\begingroup$ b) You have 3 choices for the last digit (1, 3, 5). After that, there are 5 choices for the second digits. Finally, there are 4 choices for the first digits (exclude 0 and two chosen digits). $\endgroup$ – GAVD Sep 18 '15 at 2:46
  • $\begingroup$ For (b) you can start with the units position. How many ways are there. $\endgroup$ – Shailesh Sep 18 '15 at 2:46
  • $\begingroup$ @GAVD. No. After the last digit is chosen, there are 5 ways for the first (exclude 0) and 5 for the second. So ans is 75 $\endgroup$ – Shailesh Sep 18 '15 at 2:48
  • $\begingroup$ Uhm, you are right. If second digits equals 0, then there are 5 choices for the first digits. If second digits does not equal 0, then there are 4 choices for the first digits. So, ans is $3\times 5 + 3\times 4\times 4 = 73$ $\endgroup$ – GAVD Sep 18 '15 at 2:53
  • 1
    $\begingroup$ It's best to start with last digit, then first digit and then middle. Always start with max constrained one $\endgroup$ – Shailesh Sep 18 '15 at 2:59

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!)

$$3\cdot (5\cdot 5) = 75$$

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.   $1\cdot 3\cdot 5+ 3\cdot 6 \cdot 5 = 105$

  • $\begingroup$ I really can't follow the answer for b). I don't see why my approach is wrong. We have 3 possibilities for the ones digit, I think that there should then be 6 possibilities for the tens digit (excluding the first ) and finally 4 possibilities for the hundreds digit (excluding the first , second number and 0 ). So 6*4*3=72 ? $\endgroup$ – CivilSigma Sep 18 '15 at 4:10
  • $\begingroup$ @CivilSigma You cannot exclude $0$ for the hundreds if you don't know if you've used it or not. Eg: if you pick $1$ & $0$ that leaves $\{2,3,4,5,6\}$ (no zero to exclude), but if you pick $1$ & $2$ that leaves $\{3,4,5,6\}$ (exclude the zero). $\endgroup$ – Graham Kemp Sep 18 '15 at 4:49
  • $\begingroup$ @CivilSigma, So instead: choose an odd digit for units, one of the five remaining non-zero digits for hundreds, and then one of the five remaining digits for tens. $\endgroup$ – Graham Kemp Sep 18 '15 at 4:52

I hope i am not mistaken but the answer is : in the units place you need an odd number and the only odds in the series of numbers given are 1,3,5 so its basically 3 possibilities in the tens digit place you can have any of the numbers excluding the one number that you chose for the units place so its basically either 0,1,2,3,4,6 assuming i chose 5 in the units so its 6 possibilities now in the hundreds place you have to choose out of the remaining number, you already chose a odd number that leaves you with 6 possibilities and you chose a tens digit which leaves you with 5 possibilities so its 5 the answer will be 5x6x3


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