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Math Help - [SOLVED] Number of candidates?

  1. #1
    Super Member fardeen_gen's Avatar
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    [SOLVED] Number of candidates?

    In any election, the number of contesting candidates exceeds the number of posts n for which the election is being held by r. If a voter can vote in 1980 different ways by voting for at least one candidate and can vote in 66 different ways to elect (r - 1) candidates in a similar manner, find the number of candidates losing the election and also find the value of n.
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  2. #2
    Super Member fardeen_gen's Avatar
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    Done!
    _____________________________________

    Number of posts = n
    Number of candidates = n + r
    Assume that r < n.

    A voter can vote to 1, 2, 3,...,n candidates,

    {{n+r}\choose 1} + {{n+r}\choose 2} + {{n+r}\choose 3} +\ ...\ + {{n+r}\choose n} = 1980 ---- (A)

    {{n+r}\choose 1} + {{n+r}\choose 2} + {{n+r}\choose 3} +\ ...\ + {{n+r}\choose {r-1}} = 66 ----(B)


    Rewriting LHS of (B) in reverse order and noting that,

    {{n+r}\choose{r-1}} = {{n+r}\choose{n+1}}

    {{n+r}\choose {r-2}} = {{n+r}\choose {n+2}}

    {{n+r}\choose {r-3}} = {{n+r}\choose {n+3}}
    --------------------------------
    --------------------------------

    {{n+r}\choose {n+1}} + {{n+r}\choose {n+2}} + {{n+r}\choose {n+3}} +\ ...\ + {{n+r}\choose {n+r-1}} = 66 ---- (C)

    Also note that,
    2^{n+r} = \left[{{n+r}\choose 0}\right] + \left[{{n+r}\choose 1} + {{n+r}\choose 2} + {{n+r}\choose 3} +\ ...\ + {{n+r}\choose n}\right] + \left[{{n+r}\choose {n+1}} + {{n+r}\choose {n+2}} + {{n+r}\choose {n+3}} +\ ...\ + {{n+r}\choose {n+r-1}}\right] + \left[{{n+r}\choose {n+r}}\right]

    Note that 1st box bracket = 1
    2nd box bracket = 1980 ---- {from (A)}
    3rd box bracket = 66 ---- {from (B)}
    4th box bracket = 1

    \implies\ 2^{n+r} = 1 + 1980 + 66 + 1 = 2048
    \implies\ 2^11 = 2048
    \implies\ n + r = 11

    Now solving (A) & (B) with n+r = 11, we find that,
    \boxed{n = 8\ \&\ r = 3}
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