What is expand (n(n-1)(n-2)(n-3))/(24) ?

The solution to expand (n(n-1)(n-2)(n-3))/(24) is (n^4}{24}-(n^3)/4+\frac{11n^2)/(24)-n/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

expand (n(n-1)(n-2)(n-3))/(24)

expand

\frac{n ( n - 1 ) ( n - 2 ) ( n - 3 )}{24}

\frac{n ^{4}}{24} - \frac{n ^{3}}{4} + \frac{11 n ^{2}}{24} - \frac{n}{4}

Solution steps

\frac{n ( n - 1 ) ( n - 2 ) ( n - 3 )}{24}

Expand

n (n - 1) (n - 2) (n - 3) : n^{4} - 6 n^{3} + 11 n^{2} - 6 n

= \frac{n ^{4} - 6 n ^{3} + 11 n ^{2} - 6 n}{24}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{n ^{4} - 6 n ^{3} + 11 n ^{2} - 6 n}{24} = \frac{n ^{4}}{24} - \frac{6 n ^{3}}{24} + \frac{11 n ^{2}}{24} - \frac{6 n}{24}

= \frac{n ^{4}}{24} - \frac{6 n ^{3}}{24} + \frac{11 n ^{2}}{24} - \frac{6 n}{24}

Cancel

\frac{6 n ^{3}}{24} : \frac{n ^{3}}{4}

= \frac{n ^{4}}{24} - \frac{n ^{3}}{4} + \frac{11 n ^{2}}{24} - \frac{6 n}{24}

Cancel

\frac{6 n}{24} : \frac{n}{4}

= \frac{n ^{4}}{24} - \frac{n ^{3}}{4} + \frac{11 n ^{2}}{24} - \frac{n}{4}

(x y (x, z (x)) + z^{'} (x)) (x z^{'} (x) + y (x, z (x)) z (x))

\frac{( x + 3 ) ( x - 6 )}{x - 2}

\frac{3 - x}{( x + 1 ) ( x - 1 )}

\frac{d ^{2}}{d x ^{2}} ((y) (7 x^{3} - 5 x^{2} + x))

A (- 2, - 5) y B (4, - 3)

What is expand (n(n-1)(n-2)(n-3))/(24) ?
The solution to expand (n(n-1)(n-2)(n-3))/(24) is (n^4}{24}-(n^3)/4+\frac{11n^2)/(24)-n/4