If 2- r - n- then nCr-2 nCr-1- nCr-2 is equal to-

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Binomial Coefficients

If 2≤ r ≤ n...

Question

If $2 \leq r \leq n,$ then $^{n} C_{r} + 2^{n} C_{r - 1} +^{n} C_{r - 2}$ is equal to:

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

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

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2^{n + 2} C_{r}

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^{n + 2} C_{r}

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Solution

The correct option is C $^{n + 2} C_{r}$
The value of $^{n} C_{r} + 2^{n} C_{r - 1} +^{n} C_{r - 2}$ is
$=$ $^{n} C_{r}$ + $^{n} C_{r - 1} +^{n} C_{r - 1} +^{n} C_{r - 2}$ .....Using $^{m} C_{x} +^{m} C_{x - 1} =^{m + 1} C_{x}$
$=$ $^{n + 1} C_{r}$ + $^{n + 1} C_{r - 1}$
$=$ $^{n + 2} C_{r}$

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Similar questions

Q. If

(2 \leq r \leq n),

then

^{n} C_{r} + 2 \cdot^{n} C_{r + 1} +^{n} C_{r + 2}

is equal to

$^{n} C_{r} + 2^{n} C_{r - 1} +^{n} C r - 2$ =

Q. The expression

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

is equal to

Q. If

^{n} C_{r - 1} +^{n + 1} C_{r - 1} +^{n + 2} C_{r - 1} + . . . . . . . +^{2 n} C_{r - 1}, =^{2 n + 1} C_{r^{2} - 132} -^{n} C_{r}

, then the value of

r

Let r and n be positive integers such that $1 \leq r \leq n .$ Then prove the following :

(i) $\frac{^{n} C_{r}}{^{n} C_{r - 1}} = \frac{n - r + 1}{r}$
(ii) $n^{n - 1} C_{r - 1} = (n - r + 1)^{n} C_{r - 1}$
(iii) $\frac{^{n} C_{r}}{^{n - 1} C_{r - 1}} = \frac{n}{r}$
(iv) $^{n} C_{r} + 2^{n} C_{r - 1} +^{n} C_{r - 2} =^{n + 2} C_{r}$

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If 2≤r≤n, then nCr+2nCr−1+nCr−2 is equal to:

The correct option is C n+2CrThe value of nCr+2nCr−1+nCr−2 is= nCr+nCr−1+nCr−1+nCr−2 .....Using mCx+mCx−1=m+1Cx= n+1Cr+n+1Cr−1= n+2Cr

If $2 \leq r \leq n,$ then $^{n} C_{r} + 2^{n} C_{r - 1} +^{n} C_{r - 2}$ is equal to:

The correct option is C $^{n + 2} C_{r}$
The value of $^{n} C_{r} + 2^{n} C_{r - 1} +^{n} C_{r - 2}$ is
$=$ $^{n} C_{r}$ + $^{n} C_{r - 1} +^{n} C_{r - 1} +^{n} C_{r - 2}$ .....Using $^{m} C_{x} +^{m} C_{x - 1} =^{m + 1} C_{x}$
$=$ $^{n + 1} C_{r}$ + $^{n + 1} C_{r - 1}$
$=$ $^{n + 2} C_{r}$