Quantity 2: If nC5 = nC6 , then 13Cn = ?1). Quantity 2 > Quantity 12).

1.	Quantity 2: If nC5 = nC6 , then 13Cn = ?1). Quantity 2 > Quantity 12). Quantity 2 < Quantity 13). Quantity 2 ≥ Quantity 14). Quantity 2 ≤ Quantity 1
Answer» First we will find QUANTITY 1, Quantity 1: 10 × ⁿC₂ = 3 × ^{n + 1}C₃ (? ⁿC_r= $(\frac{{n!}}{{\left( {n - r} \right)!\left( {r!} \right)}})$) $(\Rightarrow \;10\; \TIMES \;\frac{{n\left( {n - 1} \right)}}{{1\; \times \;2}}\; = \;3\; \times \;\frac{{\left( {n\; + \;1} \right)\;n\;\left( {n - 1} \right)}}{{1\; \times \;2\; \times \;3}})$ ⇒ 10 = n + 1 ⇒ n = 9 ∴ 9N = 9 × 9 = 81 Now, Quantity 2: ⁿC₅ = ⁿC₆ (? ⁿC_r = ⁿC_s then r = s or r + s = n) ⇒ n = 5 + 6 = 11 ∴ n = 11 ¹³C_n = ¹³C₁₁= ¹³C₂ $(= \;\frac{{13\; \times \;12}}{{1\; \times \;2}}\; = \;78)$ ∴ Quantity 1 > Quantity 2

Quantity 2: If nC5 = nC6 , then 13Cn = ?1). Quantity 2 > Quantity 12). Quantity 2 < Quantity 13). Quantity 2 ≥ Quantity 14). Quantity 2 ≤ Quantity 1

Answer»

First we will find QUANTITY 1,

Quantity 1:

10 × ⁿC₂ = 3 × ^{n + 1}C₃

(? ⁿC_r= $(\frac{{n!}}{{\left( {n - r} \right)!\left( {r!} \right)}})$)

$(\Rightarrow \;10\; \TIMES \;\frac{{n\left( {n - 1} \right)}}{{1\; \times \;2}}\; = \;3\; \times \;\frac{{\left( {n\; + \;1} \right)\;n\;\left( {n - 1} \right)}}{{1\; \times \;2\; \times \;3}})$

⇒ 10 = n + 1

⇒ n = 9

∴ 9N = 9 × 9 = 81

Now,

Quantity 2:

ⁿC₅ = ⁿC₆

(? ⁿC_r = ⁿC_s then r = s or r + s = n)

⇒ n = 5 + 6 = 11

∴ n = 11

¹³C_n = ¹³C₁₁= ¹³C₂ $(= \;\frac{{13\; \times \;12}}{{1\; \times \;2}}\; = \;78)$

∴ Quantity 1 > Quantity 2

Discussion

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