$^{404} C_{4} -^{4} C_{1}^{303} C_{4} +^{4} C_{2}^{202} C_{4} -^{4} C_{3}^{101} C_{4}$ is equal to

(401)^{4}

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(101)^{4}

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0

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(201)^{4}

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Solution

The correct option is B $0$
$^{404} C_{4} -^{4} C_{1} (^{303} C_{4}) +^{4} C_{2} (^{202} C_{4}) -^{4} C_{3} (^{101} C_{4})$ is equal
to
Sol $4 (^{101} C_{4}) - 4 (^{303} C_{4}) + 6 (^{202} C_{4}) - 4 (^{101} C_{4})$
$4 (^{101} C_{4}) - 4.3 (^{101} C_{4}) + 6.2 (^{101} C_{4}) - 4 (^{101} C_{4})$
$^{101} C_{4} [4 - 12 + 12 - 4]$
$^{101} C_{4} [0]$
$0$

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Similar questions

The value of $^{404} C_{4}$ - $^{4} C_{1}$ $^{303} C_{4}$ + $^{202} C_{4}$ - $^{4} C_{3}$ $^{101} C_{4}$ =

^{404} C_{4} -^{4} C_{1}^{303} C_{4} +^{4} C_{2}^{202} C_{4} -^{4} C_{3}^{101} C_{4}

is equal to

Q. If

S =^{404} C_{4} -^{4} C_{1} .^{303} C_{4} +^{4} C_{2} .^{202} C_{5} -^{4} C_{3} .^{101} C_{4} = (101)^{k}

then k equal to

\int_{0}^{4 π} c o s^{2014} x d x

is equal to

The wavelength range of the visible spectrum extends from violet (400 nm) to red (750 nm). Express these wavelength in frequencies (Hz). $(1 n m = 10^{- 9} m)$ .

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404C4−4C1303C4+4C2202C4−4C3101C4 is equal to

The correct option is B 0404C4−4C1(303C4)+4C2(202C4)−4C3(101C4) is equal to Sol 4(101C4)−4(303C4)+6(202C4)−4(101C4)4(101C4)−4.3(101C4)+6.2(101C4)−4(101C4)101C4[4−12+12−4]101C4[0]0

$^{404} C_{4} -^{4} C_{1}^{303} C_{4} +^{4} C_{2}^{202} C_{4} -^{4} C_{3}^{101} C_{4}$ is equal to