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aev [14]
2 years ago
13

How many distinct 9-letter words starting with “i” can be formed from the word “committee”?

Mathematics
1 answer:
Nikitich [7]2 years ago
7 0

We have already chosen "i" to be the first letter. This means that we still have 8 letters that we can use as the second letter. Once that letter is chosen, we still have 7 letters that we can use as the third letter, and so on. This means we have

8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 8!

ways of choosing the remaining 8 letters.

<u>However,</u> some of the 8 available letters are equal: we have 2 m's, 2 t's and 2 e's. This means that we overcounted the number of words on the reasoning above. We can correct the counting by dividing three times by 2!, one time for each of the repeated letters.

The final answer is

\dfrac{8!}{2! \times 2! \times 2!} = \dfrac{40320}{8} = 5040.

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7.4 years would be the correct answer

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The total is 275 because 100% divided by 20% is 5 and if 20%= 55 then that means that 55 goes into 100% 5 times so you multiply 55 by 5 to get 275. Hope I helped! Please rate me as the brainliest!
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Solve the recurrence relation: hn = 5hn−1 − 6hn−2 − 4hn−3 + 8hn−4 with initial values h0 = 0, h1 = 1, h2 = 1, and h3 = 2 using (
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(a) Suppose h_n=r^n is a solution for this recurrence, with r\neq0. Then

r^n=5r^{n-1}-6r^{n-2}-4r^{n-3}+8r^{n-4}
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So we expect a general solution of the form

h_n=c_1(-1)^n+(c_2+c_3n+c_4n^2)2^n

With h_0=0,h_1=1,h_2=1,h_3=2, we get four equations in four unknowns:

\begin{cases}c_1+c_2=0\\-c_1+2c_2+2c_3+2c_4=1\\c_1+4c_2+8c_3+16c_4=1\\-c_1+8c_2+24c_3+72c_4=2\end{cases}\implies c_1=-\dfrac8{27},c_2=\dfrac8{27},c_3=\dfrac7{72},c_4=-\dfrac1{24}

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(b) Let G(x)=\displaystyle\sum_{n\ge0}h_nx^n be the generating function for h_n. Multiply both sides of the recurrence by x^n and sum over all n\ge4.

\displaystyle\sum_{n\ge4}h_nx^n=5\sum_{n\ge4}h_{n-1}x^n-6\sum_{n\ge4}h_{n-2}x^n-4\sum_{n\ge4}h_{n-3}x^n+8\sum_{n\ge4}h_{n-4}x^n
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G(x)=\dfrac{x-4x^2+3x^3}{1-5x+6x^2+4x^3-8x^4}
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From here you would write each term as a power series (easy enough, since they're all geometric or derived from a geometric series), combine the series into one, and the solution to the recurrence will be the coefficient of x^n, ideally matching the solution found in part (a).
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4 0
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