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3 years ago

how many 5 digits numbers can be created using digits 1,2,3,7, and 6 without repeating any digits within that 5-digit number?

Mathematics

1 answer:

notsponge [240]3 years ago

8 0

There are 5 numbers that can be used.

The first digit can be 1 of 5

The second digit can be 1 of 4

The third digit is 1 of 3

The fourth digit is 1 of 2

The fifth digit is 1 of 1

Total digits that can be created = 5 x 4 x 3 x 2 x1 = 120

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Step-by-step explanation:

Given differential equation y''(x) − 10y'(x) + 61y(x) = −3796 cos(5x) + 185e6x

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<u>Rules for finding particular integral in some special cases:-</u>

let f(D)y = $e^{ax}$ then

the particular integral $\frac{1}{f(D)} (e^{ax} ) = \frac{1}{f(a)} (e^{ax} ) if f(a)$ ≠ 0

let f(D)y = cos (ax ) then

the particular integral $\frac{1}{f(D)} (cosax ) = \frac{1}{f(D^2)} (cosax ) =\frac{cosax}{f(-a^2)}$ f(-a^2) ≠ 0

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$(D^{2} -10D+61)y(x) = −3796 cos(5x) + 185e^{6x}$

P<u>articular integral</u>:-

$P.I = \frac{1}{f(D)}( −3796 cos(5x) + 185e^{6x})$

$P.I = \frac{1}{D^2-10D+61}( −3796 cos(5x) + 185e^{6x})$

$P.I = \frac{1}{D^2-10D+61}( −3796 cos(5x) + \frac{1}{D^2-10D+61}185e^{6x})$

P.I = $I_{1} +I_{2}$

we will apply above two conditions, we get

$I_{1}$ =

$\frac{1}{D^2-10D+61}( −3796 cos(5x) = \frac{1}{(-25)-10D+61}( −3796 cos(5x) ( since D^2 = - 5^2)$ $= \frac{1}{(36-10D}( −3796 cos(5x) \\= \frac{1}{(36-10D}X\frac{36+10D}{36+10D} ( −3796 cos(5x)$

on simplification we get

= $\frac{1}{(36^2-(10D)^2}36+10D( −3796 cos(5x)$

= $\frac{-1,36,656cos5x+1,89,800 sin5x}{1296-100(-25)}$

= $\frac{-1,36,656cos5x+1,89,800 sin5x}{-1204}$

$I_{2}$ =

$\frac{1}{D^2-10D+61}185e^{6x}) = \frac{1}{6^2-10(6)+61}185e^{6x})$

$\frac{1}{37}185e^{6x})$

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