Let n and r be two positive integers such that n - r - 2- Suppose - n- r - nCr nCr - 1 nCr - 2 n-1Cr n - 1Cr - 1 n - 1Cr - 2 n - 2Cr n - 2Cr - 1 n - 2 Cr - 2 Show that -n- r - n - 2 C3-n- 2 C3- n - 2- r - 1 Hence or otherwise-

The correct option is A (^n + 2 C_3) (^n + 1 C_3) .... (^n r + 3 C_3)/(^r + 2 C_3) (^r + 1C_3) .... (^3C_3) We know that ^mC_k = m/k #160; #160 ...

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Byju's Answer

Standard XII

Mathematics

Determinant

Let n and r b...

Question

Let n and r be two positive integers such that $n \geq r + 2$ . Suppose $Δ (n, r) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^{n} C_{r} & ^{n} C_{r + 1} & ^{n} C_{r + 2}^{n + 1} C_{r} & ^{n + 1} C_{r + 1} & ^{n + 1} C_{r + 2}^{n + 2} C_{r} & ^{n + 2} C_{r + 1} & ^{n + 2} C_{r + 2} \end{matrix} ∣ ∣ ∣ ∣ ∣$ Show that $Δ (n, r) = \frac{^{n + 2} C_{3}}{^{n + 2} C_{3}} Δ (n - 2, r - 1)$ Hence or otherwise,

\frac{(^{n + 2} C_{3}) (^{n + 1} C_{3}) . . . . (^{n - r + 3} C_{3})}{(^{r + 2} C_{3}) (^{r + 1} C_{3}) . . . . (^{3} C_{3})}

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- \frac{(^{n + 2} C_{3}) (^{n + 1} C_{3}) . . . . (^{n - r + 3} C_{3})}{(^{r + 2} C_{3}) (^{r + 1} C_{3}) . . . . (^{3} C_{3})}

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\frac{(^{n + 3} C_{3}) (^{n + 2} C_{3}) . . . . (^{n - r + 3} C_{3})}{(^{r + 3} C_{3}) (^{r + 2} C_{3}) . . . . (^{3} C_{3})}

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- \frac{(^{n + 3} C_{3}) (^{n + 2} C_{3}) . . . . (^{n - r + 3} C_{3})}{(^{r + 3} C_{3}) (^{r + 2} C_{3}) . . . . (^{3} C_{3})}

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Solution

The correct option is A $\frac{(^{n + 2} C_{3}) (^{n + 1} C_{3}) . . . . (^{n - r + 3} C_{3})}{(^{r + 2} C_{3}) (^{r + 1} C_{3}) . . . . (^{3} C_{3})}$
We know that $^{m} C_{k} = \frac{m}{k} .^{m - 1} C_{k - 1}$ . Therefore we can write
$Δ (n, r) = ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{n}{r}^{n - 1} C_{r - 1} & \frac{n}{(r + 1)}^{n - 1} C_{r} & \frac{n}{(r + 2)}^{n - 1} C_{r + 1} \frac{n + 1}{r}^{n} C_{r - 1} & \frac{(n + 1)}{(r + 1)}^{n} C_{r} & \frac{(n + 1)}{(r + 2)}^{n} C_{r + 1} \frac{n + 2}{r}^{n + 1} C_{r - 1} & \frac{(n + 2)}{(r + 1)}^{n + 1} C_{r} & \frac{(n + 2)}{(r + 2)}^{n + 1} C_{r + 1} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣$

$∴ Δ (n, r) = \frac{n (n + 1) (n + 2)}{r (r + 1) (r + 2)} ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^{n - 1} C_{r - 1} & ^{n - 1} C_{r} & ^{n - 1} C_{r + 1}^{n} C_{r - 1} & ^{n} C_{r} & ^{n} C_{r + 1}^{n + 1} C_{r - 1} & ^{n + 1} C_{r} & ^{n + 1} C_{r + 1} \end{matrix} ∣ ∣ ∣ ∣ ∣$
$Δ (n, r) = \frac{^{n + 2} C_{3}}{^{r + 2} C_{3}} Δ (n - 1, r - 1)$
$= \frac{^{n + 2} C_{3}}{^{r + 2} C_{3}} \cdot \frac{^{n + 1} C_{3}}{^{r + 1} C_{3}} \cdot \frac{^{n} C_{3}}{^{r} C_{3}} \cdot Δ (n - 3, r - 3)$
$= \frac{(^{n + 2} C_{3}) (^{n + 1} C_{3}) (^{n} C_{3}) . . . . (^{n - r + 3} C_{3})}{(^{r + 2} C_{3}) (^{r + 1} C_{3}) (^{r} C_{3}) . . . . . (^{3} C_{3})} Δ (n - r, 0)$ (1)
Now $Δ (n - r, 0) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^{n - r} C_{0} & ^{n - r} C_{1} & ^{n - r} C_{2}^{n - r + 1} C_{0} & ^{n - r + 1} C_{1} & ^{n - r + 1} C_{2}^{n - r + 2} C_{0} & ^{n - r + 2} C_{1} & ^{n - r + 2} C_{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$
$= ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & (n - r) & \frac{(n - r) (n - r - 1)}{2} 1 & (n - r + 1) & \frac{(n - r + 1) (n - r)}{2} 1 & (n - r + 2) & \frac{(n - r + 2) (n - r + 1)}{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$
Applying $R_{2} \to R_{2} - R_{1}; R_{3} \to R_{3} - R_{2}$
$= ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & n - r & \frac{(n - r) (n - r - 1)}{2} 0 & 1 & n - r 0 & 1 & n - r + 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣$
$= 1$
Hence $Δ (n, r) = \frac{(^{n + 2} C_{3}) (^{n + 1} C_{3}) . . . . (^{n - r + 3} C_{3})}{(^{r + 2} C_{3}) (^{r + 1} C_{3}) . . . . (^{3} C_{3})}$

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Similar questions

Q. Let r and n be positive integers such that 1 ≤ r ≤ n. Then prove the following:
(a)

\frac{{}^{n}C_{r}}{{}^{n}C_{r - 1}} = \frac{n - r + 1}{r}

(b) n ·_{^{n −}}_¹C_r_{− 1} = (n − r + 1) ⁿC_r_{− 1}
(c)

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

(iv) ⁿC_r + 2 · ⁿC_r_{− 1} + ⁿC_r_{− 2} = ^{n +}²C_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) $^{n} C_{r} + 2^{n} C_{r - 1} +^{n} C_{r - 2} =^{n + 2} C_{r}$

Q. Prove that :

C_{1} + 2 C_{2} + 3 C_{3} + . . n C_{n} = n {.2}^{n - 2}

Q. Prove that

C_{1} + 2 C_{2} + 3 C_{3} + 4 C_{4} + . +^{n} C_{n} = n {.2}^{n - 1}

Q. Assertion :For

n \in N

, let

C_{r} = (\begin{matrix} n r \end{matrix})

0 \leq r \leq n

\sum_{r = 1}^{n} (r) C_{r}^{2} = n (\begin{matrix} 2 n - 1 n - 1 \end{matrix})

Reason:

\sum_{r = 1}^{n} C_{r}^{2} = (\begin{matrix} 2 n n \end{matrix})

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Let n and r be two positive integers such that n≥r+2. Suppose Δ(n,r)=∣∣ ∣ ∣∣nCrnCr+1nCr+2n+1Crn+1Cr+1n+1Cr+2n+2Crn+2Cr+1n+2Cr+2∣∣ ∣ ∣∣ Show that Δ(n,r)=n+2C3n+2C3Δ(n−2,r−1) Hence or otherwise,