[SOLVED] Number of candidates?

**fardeen_gen** · May 29th 2009, 08:04 PM

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.

**fardeen_gen** · June 3rd 2009, 10:09 AM

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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