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KEAM 2011 Math Paper

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Question : 97 of 120

Marks: +1, -0

If

{}^{n}C_{r-1}=28,{}^{n}C_{r}=56

{}^{n}C_{r+1}=70,

then the value of

r

is equal to

1
2
3
4
5

Solution:

{}^{n}C_{r-1}=28,{}^{n}C_{r}=56.{}^{n}C_{r+1}=70

\Rightarrow \frac{{}^{n}C_{r}}{{}^{n}C_{r-1}}=\frac{56}{28}

\Rightarrow \frac{ \frac{n!}{r(r-1)!(n-r)!} }{ \frac{n!}{(r-1)!(n-r+1)(n-r)!} } = 2

=n-r+1=2r

\Rightarrow n-3r=-1

.....(i)

\frac{{}^{n}C_{r+1}}{{}^{n}C_{r}} = \frac{70}{56}

\Rightarrow \frac{ \frac{n!}{(r+1)!(n-r-1)!} }{ \frac{n!}{r!(n-r)!} } = \frac{35}{28}

\Rightarrow \frac{ \frac{n!}{(r+1)r!(n-r-1)!} }{ \frac{n!}{r!(n-r)(n-r-1)!} } = \frac{35}{28}

\Rightarrow \frac{n-r}{r+1} = \frac{35}{28}

\Rightarrow 28n-28r=35r+35

\Rightarrow 28n-63r=35

....(ii)

Multiply by 21 in Eq. (i) and subtracting from Eq. (ii),

21n-63r=-21

28n-63r=35

-

+

-

----------------------

-7n=-56 \Rightarrow n=8

From Eq. (i),

3r=n+1=8+1

r=3

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