mathmath333
  • mathmath333
Counting Problem
Mathematics
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mathmath333
  • mathmath333
Counting Problem
Mathematics
chestercat
  • chestercat
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mathmath333
  • 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
  • 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
  • mathmath333
at b, there can be remaining 5 numbers where does 5 come from ???

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hartnn
  • 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
  • mathmath333
but even numbers are three (2,4,6) not one
hartnn
  • 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
  • mathmath333
yes
hartnn
  • 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
  • mathmath333
but why are u starting/counting from unit's digit rather than hundred's digit
hartnn
  • hartnn
because there is restriction at unit's place and not at hundred's place
mathmath333
  • mathmath333
answer is 60 (given)
hartnn
  • 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
  • mathmath333
ohk i see ur logic
ganeshie8
  • 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
  • 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
  • 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
  • ganeshie8
Haha exactly ^
terenzreignz
  • terenzreignz
Sorry... I keep missing that there's no 5 in the set. Dang misleading questions...
terenzreignz
  • terenzreignz
Though clever...
ganeshie8
  • 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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