Find the value of x such that: $^{7}$ C $_{x - 1}$ + $^{7}$ C $_{x}$ = $^{8}$ C $_{x + 2}$

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Solution

The correct option is B 3
Soln: We have $^{n}$ C $_{r - 1}$ + $^{n}$ C $_{r}$ = n + $^{1}$ C $_{r}$

$^{7}$ C $_{x - 1}$ + $^{7}$ C $_{x}$ = $^{8}$ C $_{x}$

$^{8}$ C $_{x}$ = $^{8}$ C $_{x + 2}$ = $^{8}$ C $_{(8 - x - 2)}$

Either x = x + 2 or x + 2 = (8 - x - 2)

But x $\neq$ x+2

2x + 2 = 8

x = 3

Hence option (b)

Alternatively, go from answer options.

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^{7} C_{x - 1} +^{7} C_{x} =^{8} C_{x + 2}

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