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3 years ago

5

Combinations- how to show that 10C2=10C8

1 answer:

Black_prince [1.1K]3 years ago

7 0

By definition,

${}^nC_k=\dbinom nk=\dfrac{n!}{k!(n-k)!}$

We have

$\dbinom{10}2=\dfrac{10!}{2!(10-2)!}=\dfrac{10!}{2!8!}=\dfrac{10!}{(10-8)!8!}=\dbinom{10}8$

and so

${}^{10}C_8={}^{10}C_2$

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3 years ago

Will crown the correct answers brainliest 3x

Solnce55 [7]

Answer :

(1) $\frac{1}{x}+\frac{1}{y}=\frac{y+x}{xy}$

(2) $\frac{1}{5x}+\frac{1}{3y}=\frac{3y+5x}{15xy}$

(3) $\frac{1}{7x}-\frac{1}{2y}=\frac{2y-7x}{14xy}$

(4) $\frac{2}{5x}+\frac{3}{7y}=\frac{6y+15x}{21xy}$

(5) $\frac{7}{11x}-\frac{1}{33y}=\frac{231y-11x}{363xy}$

(6) $\frac{1}{2x}+\frac{3}{x}-\frac{1}{7y}=\frac{7y+42y-2x}{14xy}$

Step-by-step explanation :

(1) The given expression is: $\frac{1}{x}+\frac{1}{y}$

$\frac{1}{x}+\frac{1}{y}=\frac{y+x}{xy}$

(2) The given expression is: $\frac{1}{5x}+\frac{1}{3y}$

$\frac{1}{5x}+\frac{1}{3y}=\frac{3y+5x}{(5x)\times (3y)}=\frac{3y+5x}{15xy}$

(3) The given expression is: $\frac{1}{7x}-\frac{1}{2y}$

$\frac{1}{7x}-\frac{1}{2y}=\frac{2y-7x}{(7x)\times (2y)}=\frac{2y-7x}{14xy}$

(4) The given expression is: $\frac{2}{5x}+\frac{3}{7y}$

$\frac{2}{5x}+\frac{3}{7y}=\frac{(2\times 3y)+(3\times 5x)}{(5x)\times (7y)}=\frac{6y+15x}{21xy}$

(5) The given expression is: $\frac{7}{11x}-\frac{1}{33y}$

$\frac{7}{11x}-\frac{1}{33y}=\frac{(7\times 33y)-(1\times 11x)}{(11x)\times (33y)}=\frac{231y-11x}{363xy}$

(6) The given expression is: $\frac{1}{2x}+\frac{3}{x}-\frac{1}{7y}$

$\frac{1}{2x}+\frac{3}{x}-\frac{1}{7y}=\frac{(x\times 7y)+(3\times 2x\times 7y)-(2x\times x)}{(2x)\times (x)\times (7y)}=\frac{7xy+42xy-2x^2}{14x^2y}=\frac{7y+42y-2x}{14xy}$

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Step-by-step explanation:

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3 years ago

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