Permutation: President and Vice-President

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Twenty eight.

You can make $30$ choices of a President and Vice-President of $6$ men: any one of the $6$ can be elected as a President, then any of the remaining 5 as a Vice-President, and that makes $6\cdot 5=30$.
But two of possible results, Carl+Fred and Fred+Carl, are forbidden, hence the answer $30-2 = 28$.

Or more generally: there is $\frac{n!}{(n-k)!}$ k-permutations of n, so in your problem there is $\frac{6!}{(6-2)!}$ possible choices, $\frac{2!}{(2-2)!}$ of which are forbidden, giving a result of $6\cdot 5 - 2=28$.

You may also describe it with specific sequences of events: first elect a President, which makes $6$ possibilities; then if a President is Carl or Fred, choose one of four for a Vice-President, otherwise choose on of five.
Finally there are $2\cdot 4 + 4\cdot 5 = 28$ outcomes.

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Updated on January 14, 2020

Comments

  • MC UO
    MC UO almost 4 years

    Al, Bert, Carl, Don, Ernie and Fred are $6$ men in a club that is electing a President and Vice-President. In how many ways can this be done if Carl and Fred cannot be together as President an Vice-President?

    My answer is: $480$ ways

    • Em.
      Em. almost 8 years
      You should include in your posts your thought process, not just a numerical value.
  • N. F. Taussig
    N. F. Taussig almost 8 years
    You could have stopped after Case 2.