MATHS GENIUS NEEDED URGENT....

find the number of odd integers between 30000 and 80000 in which no digit is repeated.

Odd means that the last digit is either 1, 3, 5, 7, or 9

So we have 5 choices, for the last digit.
We have to be a little careful, since the first digit can only range from 3 to 7 (largest possible number is 79,999)

So if we choose 1 or 9, then we can choose ANY of those number for the first digit. But if we choose 3, 5, or 7, then we need to take one out of those possible numbers:

In all we have:

xx,xxx = 5 digits

If we choose 1 or 9, then we can choose any of the 3 - 7 = 5 digits (3, 4, 5, 6, 7) for the first digit...

So this is 2 * 5 = 10 different ways to choose 1 or 9 as the last digit and 3-7 as the first....now just choose the last three (they can be ANY digit):

We have already picked out 2 numbers, this only leaves:

10 - 2 = 8, then 7, then 6 numbers to choose from:

So if 1 or 9 is the last digit, then there are:

2*5*8*7*6 = 3,360
(notice there were FIVE multiplications for EACH place in the FIVE digit number)

Now if we choose 3, 5 or 7, then we can only CHOOSE 4 of the 5 valid values for the first digit (because one will be taken out)...again that leaves 8 then 7 then 6 digits for the three middle places:

3*4*8*7*6 = 4,032

Now just add those two up to get the total odd numbers ending with 1 or 9 OR ending with 3, 5, or 7:

3,360 + 4,032 = 7,392


Edit:

Perhaps a few examples will make it clearer:

Choose 1 as the last digit and choose 3 as the first:

3x,xx1
--> you can put 2, then 4, then 5 in

32,451
or
32,461
32,471
32,481
32,491
32,401

So once you choose 1, 3, 2 and 4, there are SIX possibilities for the third digit (so nothing is repeated).

If we ONLY choose 1, 3, and 2, then we can change 4 into SEVEN different values:

1) 32,4x1
2) 32,5x1