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Question:

(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?

My Solution:

a)(7)(6)(5).

Don't know the answers of other 2.

Correct me if I am wrong. Thanks!!!

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There are two possible answers to this depending of if the number can start with a 0 or not.

If it can then the number is $7 \times 6 \times 5 = 210$

However we don't usually write numbers with leading zeros. So assuming we don't allow a leading zero the answer is $6 \times 6 \times 5 = 180$

How many of these are odd?

Lets consider picking the numbers in a special order

First we need to pick a 1, 3 or 5 so in our first draw for the final digit so we have a choice of 3. Next we pick our first digit which can be any thing apart from 0 or the number we have just picked making a choice of 5 and for our final draw the middle digit we can pick any of the 5 remaining digits making $3 \times 5 \times 5 = 75$

Finally For the third one greater than 330 we have two ways to achieve this either draw a 4, 5 or 6 for the first digit then we don't care about the others making $3 \times 6 \times 5 = 90$

Alternatively we must draw a 3 for our first digit then 4, 5 or 6 for our second and finally any digit making $1 \times 3 \times 5 = 15$

Now since both of these groups have no overlap we can simply add them together to get $90 + 15 = 105$.

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It has to be $180$.

$6$ choices for the hundreds position as out of $7$ given numbers, we have to forget zero otherwise it shall become a two digit number. Since we have consumed one number and now zero can be used so for the tens position we can use 6 numbers. Now for unit position we have left with $5$ choices of numbers. So $6\times 6\times 5=180$.

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Any of the 6 nonzero digits can be chosen for the hundreds position, and of the remaining 6 digits for the tens position, leaving 5 digits for the units position. So, there are (6)(6)(5) = 180 three digit numbers.

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  • $\begingroup$ It should be $6\times 6\times 5$ as you and other post suggested long ago. $\endgroup$
    – user99914
    Oct 3, 2015 at 9:53
  • $\begingroup$ I 110% agree with you, and till now I don't understand why the solution manual wrote it like this (the answer I wrote is copied from the solution manual !!!!!! ) .... stupid solution manual -_- :D $\endgroup$
    – Ammar Rashed
    Oct 3, 2015 at 10:04
  • $\begingroup$ However that can be true if ' 0 ' is to be used :D $\endgroup$
    – Ammar Rashed
    Oct 3, 2015 at 10:05

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