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mathmath333

  • one year ago

Counting Problem

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \normalsize \text{How many 3-digit even numbers can be made using the digits}\hspace{.33em}\\~\\ & \normalsize \text{1, 2, 3, 4, 6, 7}\quad \text{ if no digit is repeated?}\hspace{.33em}\\~\\ \end{align}}\)

  2. hartnn
    • one year ago
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    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
    • one year ago
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    at b, there can be remaining 5 numbers where does 5 come from ???

  4. hartnn
    • one year ago
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    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
    • one year ago
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    but even numbers are three (2,4,6) not one

  6. hartnn
    • one year ago
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    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
    • one year ago
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    yes

  8. hartnn
    • one year ago
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    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
    • one year ago
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    but why are u starting/counting from unit's digit rather than hundred's digit

  10. hartnn
    • one year ago
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    because there is restriction at unit's place and not at hundred's place

  11. mathmath333
    • one year ago
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    answer is 60 (given)

  12. hartnn
    • one year ago
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    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
    • one year ago
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    ohk i see ur logic

  14. ganeshie8
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    Haha exactly ^

  18. terenzreignz
    • one year ago
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    Sorry... I keep missing that there's no 5 in the set. Dang misleading questions...

  19. terenzreignz
    • one year ago
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    Though clever...

  20. ganeshie8
    • one year ago
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    the given set has some sort of symmetry with respect to even/odd which we can use to our advantage :)

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