If $^{n} C_{r} +^{n} C_{r + 1} =^{n + 1} C_{x}$ then the value of $x$ is equal to

r

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r + 1

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r - 1

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2 r

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Solution

The correct option is A $r + 1$
Given,
$^{n} C_{r} +^{n} C_{r + 1} =^{n + 1} C_{x}$
or, $^{n} C_{r + 1} +^{n} C_{r + 1 - 1} =^{n + 1} C_{x}$
or, $^{n + 1} C_{r + 1} =^{n + 1} C_{x}$
or, $x = r + 1$ . [ Comparing both sides we have]

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Similar questions

If $^{n} C_{r} +^{n} C_{r + 1} =^{n + 1} C_{x},$ then x =

^{n} C_{r} +^{n} C_{r + 1} =^{(n + 1)} C_{x}

x =

^{n} P_{r} =^{n} P_{r + 1}

^{n} C_{r} =^{n} C_{r - 1}

r

2 \leq r \leq n,

^{n} C_{r} + 2^{n} C_{r - 1} +^{n} C_{r - 2}

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