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Oduvanchick [21]
3 years ago
11

Suppose a company needs temporary passwords for the trial of a new payroll software. Each password will have one digit followed

by three letters. The letter and the digits , , , and will not be used. So, there are letters and digits that will be used. Assume that the letters can be repeated. How many passwords can be created using this format?
Mathematics
1 answer:
Alekssandra [29.7K]3 years ago
3 0

Answer:  The required number of passwords that can be created is 175760.

Step-by-step explanation:  Given that a company needs temporary passwords for the trial of a new payroll software.

Each password will have one digit followed by three letters and the letters can be repeated.

We are to find the number of passwords that can be created using this format.

For the one digit in the password, we have 10 options, 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9.

Since there are 26 letters in English alphabet and letters can be repeated, so the number of options for 3 letters is

26 × 26 × 26 = 17576.

Therefore, the total number of ways in which passwords can be created using the given format is

n=9\times17576=175760.

Thus, the required number of passwords that can be created is 175760.

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Find a compact form for generating functions of the sequence 1, 8,27,... , k^3
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This sequence has generating function

F(x)=\displaystyle\sum_{k\ge0}k^3x^k

(if we include k=0 for a moment)

Recall that for |x|, we have

\displaystyle\frac1{1-x}=\sum_{k\ge0}x^k

Take the derivative to get

\displaystyle\frac1{(1-x)^2}=\sum_{k\ge0}kx^{k-1}=\frac1x\sum_{k\ge0}kx^k

\implies\dfrac x{(1-x)^2}=\displaystyle\sum_{k\ge0}kx^k

Take the derivative again:

\displaystyle\frac{(1-x)^2+2x(1-x)}{(1-x)^4}=\sum_{k\ge0}k^2x^{k-1}=\frac1x\sum_{k\ge0}k^2x^k

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Take the derivative one more time:

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\implies\displaystyle\frac{x+4x^3+x^3}{(1-x)^4}=\sum_{k\ge0}k^3x^k

so we have

\boxed{F(x)=\dfrac{x+4x^3+x^3}{(1-x)^4}}

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