## mathmath333 one year ago Counting Problem

1. mathmath333

\large \color{black}{\begin{align} & \normalsize \text{How many 3-digit even numbers can be made using the digits}\hspace{.33em}\\~\\ & \normalsize \text{1, 2, 3, 4, 6, 7}\quad \text{ if no digit is repeated?}\hspace{.33em}\\~\\ \end{align}}

2. hartnn

let the number be abc, at 'c', there can be one of the 3 even numbers. at b, there can be remaining 5 numbers at a, there can be remaining 4 numbers. total numbers = $$3\times 5 \times 4$$

3. mathmath333

at b, there can be remaining 5 numbers where does 5 come from ???

4. hartnn

out of these 6 numbers, 1,2,3,4,6,7 one is used in the units place, so there are 5 numbers to choose from.

5. mathmath333

but even numbers are three (2,4,6) not one

6. hartnn

yes, thats why '3' is multiplied, not one. but you will use only one of the 3 even numbers in the unit's place right?

7. mathmath333

yes

8. hartnn

if we could repeat the numbers, then we would have got 6 options at both a and b position, then it would be 6*6*3

9. mathmath333

but why are u starting/counting from unit's digit rather than hundred's digit

10. hartnn

because there is restriction at unit's place and not at hundred's place

11. mathmath333

12. hartnn

if we say there are 6 options at hundreds place, that would be incorrect, because one of them, an even number, must be used at unit's place.

13. mathmath333

ohk i see ur logic

14. ganeshie8

Alternatively, you may see that total possible 3 digit integers with distinct digits is $$6*5*4 = 120$$. However exactly half of these odd, so $$120/2 = 60$$ are even.

15. terenzreignz

You lost me... long ago. But I'm back But you also lost me at "exactly half of these are odd" Really? :D

16. hartnn

we can say exactly half of these are odd, only because half of the given digits are odd! otherwise, that statement won't hold true.

17. ganeshie8

Haha exactly ^

18. terenzreignz

Sorry... I keep missing that there's no 5 in the set. Dang misleading questions...

19. terenzreignz

Though clever...

20. ganeshie8

the given set has some sort of symmetry with respect to even/odd which we can use to our advantage :)