n C r -2 n C r -1- n Cr -2-A- n-1 CrB- n-1 Cr-1C- n-2 CrD- n-2 Cr-1

The correct option is C ^nC_r + 2^nC_r 1 + ^nC_r 2 = ^nC_r + ^nC_r 1 +^nC_r 1 + nbsp;^nC_r 2 = ^n+1C_r + ^n+1C_r 1 = ^n+2C_r ...

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Byju's Answer

Standard XI

Mathematics

Binomial Coefficients

n C r +2 n C ...

Question

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

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Solution

The correct option is C

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

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

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Similar questions

Q. The expression

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

is equal to

Q. Find the value of

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

$^{n} C_{r} +^{n} C_{r - 1}$ is equal to

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

Q. If

(2 \leq r \leq n),

then

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

is equal to

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nCr+2nCr−1+nCr−2 =

The correct option is C nCr+2nCr−1+nCr−2 = nCr+nCr−1+nCr−1+nCr−2 = n+1Cr+n+1Cr−1 = n+2Cr

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

The correct option is C

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

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