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3241004551 [841]
3 years ago
5

Two n-digit sequences of digits 0,1,. . . ,9 are said to be of the same type if the digits of one are a permutation of the digit

s of the other. For n = 8, for example, the sequences 03088929 and 00238899 are the same type. How many types of n-digit sequences are there?
Mathematics
1 answer:
Stells [14]3 years ago
6 0

Answer:

(n+9)!/(n!9!)

Step-by-step explanation:

A combination is a selection of items from a collection, such that (unlike permutations) the order of selection does not matter.

This is a case of combination with repetition because any n-digit sequence can have numbers repeated. For example in the sequence 00238899  there are 2 zeros.

The formula to find the number of combinations when repetition is allowed is:

(n+N-1)!/[n!(N-1)!]

where

N is the number of elements of the set we can choose from

n is the number of elements we choose

In this case, the set {0,1,2,3,4,5,6,7,8,9} has 10 elements, N=10

The number of n-digit sequences is given by

(n+10-1)!/[n!(10-1)!]=(n+9)!/(n!9!)

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Answer:

The vertex: (\frac{3}{4},-\frac{41}{4} )

The vertical intercept is: y=-8

The coordinates of the two intercepts of the parabola are (\frac{3+\sqrt{41} }{4} , 0) and (\frac{3-\sqrt{41} }{4} , 0)

Step-by-step explanation:

To find the vertex of the parabola 4x^2-6x-8 you need to:

1. Find the coefficients <em>a</em>, <em>b</em>, and <em>c </em>of the parabola equation

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x=-\frac{b}{2a}, so

x=-\frac{-6}{2\cdot 4}\\x=\frac{3}{4}

3. To find the y-coordinate of the vertex you use the parabola equation and x-coordinate of the vertex (f(-\frac{b}{2a})=a(-\frac{b}{2a})^2+b(-\frac{b}{2a})+c)

f(-\frac{b}{2a})=a(-\frac{b}{2a})^2+b(-\frac{b}{2a})+c\\f(\frac{3}{4})=4\cdot (\frac{3}{4})^2-6\cdot (\frac{3}{4})-8\\y=\frac{-41}{4}

To find the vertical intercept you need to evaluate x = 0 into the parabola equation

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To find the coordinates of the two intercepts of the parabola you need to solve the parabola by completing the square

\mathrm{Add\:}8\mathrm{\:to\:both\:sides}

x^2-6x-8+8=0+8

\mathrm{Simplify}

4x^2-6x=8

\mathrm{Divide\:both\:sides\:by\:}4

\frac{4x^2-6x}{4}=\frac{8}{4}\\x^2-\frac{3x}{2}=2

\mathrm{Write\:equation\:in\:the\:form:\:\:}x^2+2ax+a^2=\left(x+a\right)^2

x^2-\frac{3x}{2}+\left(-\frac{3}{4}\right)^2=2+\left(-\frac{3}{4}\right)^2\\x^2-\frac{3x}{2}+\left(-\frac{3}{4}\right)^2=\frac{41}{16}

\left(x-\frac{3}{4}\right)^2=\frac{41}{16}

\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}

x_1=\frac{\sqrt{41}+3}{4},\:x_2=\frac{-\sqrt{41}+3}{4}

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