The expression $^{n + 4} C_{r} -^{n} C_{r} - {3.}^{n} C_{r - 1} - 3^{n} C_{r - 2} -^{n} C_{r - 3}$

^{n + 3} C_{r - 1}

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

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^{n + 4} C_{r + 1}

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

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Solution

The correct option is A $^{n + 3} C_{r - 1}$
$^{n + 4} C_{r}$ - $^{n} C_{r}$ -3. $^{n} C_{r - 1}$ -3. $^{n} C_{r - 2}$ - $^{n} C_{r - 3}$
$⟹$ $^{n + 4} C_{r}$ -(( $^{n} C_{r}$ + $^{n} C_{r - 1}$ )+2.( $^{n} C_{r - 1}$ + $^{n} C_{r - 2}$ )+( $^{n} C_{r - 2}$ + $^{n} C_{r - 3})$
$⟹$ $^{n + 4} C_{r}$ -( $^{n + 1} C_{r}$ +2. $^{n + 1} C_{r - 1}$ + $^{n + 1} C_{r - 2}$ )
$⟹$ $^{n + 4} C_{r}$ -(( $^{n + 1} C_{r}$ + $^{n + 1} C_{r - 1}$ )+( $^{n + 1} C_{r - 1}$ + $^{n + 1} C_{r - 2}$ ))
$⟹$ $^{n + 4} C_{r}$ -( $^{n + 2} C_{r}$ + $^{n + 2} C_{r - 1}$ )
$⟹$ $^{n + 3} C_{r}$ + $^{n + 3} C_{r - 1}$ - $^{n + 3} C_{r}$
$⟹$ $^{n + 3} C_{r - 1}$

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Similar questions

Q. The expression

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

is equal to

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

Q. If

(2 \leq r \leq n),

then

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

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. Prove that

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

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The expression n+4Cr−nCr−3.nCr−1−3nCr−2−nCr−3

The correct option is A n+3Cr−1n+4Cr-nCr-3.nCr−1-3.nCr−2-nCr−3⟹n+4Cr-((nCr+nCr−1)+2.(nCr−1+nCr−2)+(nCr−2+nCr−3)⟹n+4Cr-(n+1Cr+2.n+1Cr−1+n+1Cr−2)⟹n+4Cr-((n+1Cr+n+1Cr−1)+(n+1Cr−1+n+1Cr−2))⟹n+4Cr-(n+2Cr+n+2Cr−1)⟹n+3Cr+n+3Cr−1-n+3Cr⟹n+3Cr−1

The expression $^{n + 4} C_{r} -^{n} C_{r} - {3.}^{n} C_{r - 1} - 3^{n} C_{r - 2} -^{n} C_{r - 3}$