What is the value of $^{10} C_{8} +^{10} C_{2}$ ?

45

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100

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90

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50

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Solution

The correct option is C $90$
We know that $^{n} C_{r} = \frac{n!}{(n - r)! \times r!} .$

Solving the first term $^{10} C_{8}$ , where $n = 10$ and $r = 8$ we get,
$^{10} C_{8} = \frac{10!}{(10 - 8)! \times 8!}$

$\Rightarrow^{10} C_{8} = \frac{10!}{2! \times 8!}$

$\Rightarrow^{10} C_{8} = \frac{10 \times 9 \times 8!}{2 \times 1 \times 8!}$

$\Rightarrow^{10} C_{8} = \frac{10 \times 9}{2} = \frac{90}{2} = 45$

Similarly,
$^{10} C_{2} = \frac{10!}{(10 - 2)! \times 2!}$

$\Rightarrow^{10} C_{2} = \frac{10!}{8! \times 2!}$

It has same value as above term. So, value of $^{10} C_{2} = 45$

Now, adding both terms we get,
$^{10} C_{8} +^{10} C_{2} = 45 + 45 = 90$

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Similar questions

^{10} C_{1} +^{10} C_{2} +^{10} C_{3} + . . . +^{10} C_{9}

(21 C_{1} - 10 C_{1}) + (^{21} C_{2} - 10 C_{2}) + (21 C_{3} - 10 C_{3}) + (21 C_{4} - 10 C_{4}) + \dots . + (21 C_{10} - 10 c_{10})

(^{21} C_{1} -^{10} C_{1}) + (^{21} C_{2} -^{10} C_{2}) + (^{21} C_{3} -^{10} C_{3}) + (^{21} C_{4} -^{10} C_{4}) + \dots + (^{21} C_{10} -^{10} C_{10})

x^{2} + p x + 45 = 0

^{10} C_{2}

p

56 c - 10 c^{2}

10 c^{2}

(- 96 c^{2} - 23 c)

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