Counting Problem

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- mathmath333

Counting Problem

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- mathmath333

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

- 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 = \(3\times 5 \times 4 \)

- mathmath333

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

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## More answers

- 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.

- mathmath333

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

- 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?

- mathmath333

yes

- 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

- mathmath333

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

- hartnn

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

- mathmath333

answer is 60 (given)

- 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.

- mathmath333

ohk i see ur logic

- 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.

- terenzreignz

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

- 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.

- ganeshie8

Haha exactly ^

- terenzreignz

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

- terenzreignz

Though clever...

- ganeshie8

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

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