The value of n Cr-2n Cr-1-n Cr-2- 2 - r - n- is

NCr + 2nCr minus;1 + nCr minus;2 nbsp; = nbsp;nCr nbsp;+ nbsp;nCr minus;1 nbsp;+ nCr minus;1 nbsp;+ nCr minus;2 nbsp; = n+1Cr + n+1Cr ...

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Standard VIII

Mathematics

Expansion of (x+y)^2

The value of ...

Question

The value of ⁿC_r + 2 ⁿC_r _{– 1} + ⁿC_r _{– 2}, 2 ≤ r ≤ n, is ____________.

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Solution

ⁿC_r + 2ⁿC_r−1 + ⁿC_r−2
= ⁿC_r + ⁿC_r−1 + ⁿC_r−1+ ⁿC_r−2
= ⁿ⁺¹C_r + ⁿ⁺¹C_r−1
= ⁿ⁺²C_r using identity
∴ ⁿC_r +2 ⁿC_r−1+ ⁿC_r−2 = ⁿ⁺²C₂

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Q. If

2 \leq r \leq n,

then

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

is equal to:

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

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

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

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

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The value of nCr + 2 nCr – 1 + nCr – 2, 2 ≤ r ≤ n, is ____________.

nCr + 2nCr−1 + nCr−2 = nCr + nCr−1 + nCr−1 + nCr−2 = n+1Cr + n+1Cr−1 = n+2Cr using identity ∴ nCr + 2 nCr−1 + nCr−2 = n+2C2

The value of ⁿC_r + 2 ⁿC_r _{– 1} + ⁿC_r _{– 2}, 2 ≤ r ≤ n, is ____________.

ⁿC_r + 2ⁿC_r−1 + ⁿC_r−2
= ⁿC_r + ⁿC_r−1 + ⁿC_r−1+ ⁿC_r−2
= ⁿ⁺¹C_r + ⁿ⁺¹C_r−1
= ⁿ⁺²C_r using identity
∴ ⁿC_r +2 ⁿC_r−1+ ⁿC_r−2 = ⁿ⁺²C₂