Simple counting exercise. Students are encouraged to look for the smart way of counting.

(a) The number of words is $(\\var{c} + \\var{v})^\\var{w} = \\var{c+v}^\\var{w}$.

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(b) The number of $\\var{w}$-letter words with exactly $n$ vowels is

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(number of ways of choosing positions of the vowels) $\\times$ (number of ways of choosing the vowels) $\\times$ (number of ways of choosing the consonants) = $\\begin{pmatrix} \\var{w} \\\\ n \\end{pmatrix} \\times \\var{v}^n \\times \\var{c}^{(\\var{w}-n)}$.

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The total is

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$\\displaystyle \\sum\\limits_{n=0}^\\var{w} \\left( \\begin{pmatrix} \\var{w} \\\\ n \\end{pmatrix} \\times \\var{v}^n \\times \\var{c}^{(\\var{w}-n)} \\right) = (\\var{v} + \\var{c})^\\var{w}$.

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(c) The number of words with at least one vowel is

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number of words with exactly one vowel + number of words with exactly two vowels + number of words with exactly three vowels + number of words with exactly four vowels + number of words with exactly five vowels = $\\var{answers[1]} + \\var{answers[2]} +\\var{answers[3]} +\\var{answers[4]} +\\var{answers[5]} = \\var{ans3}$.

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Alternatively, it is all the words except those with no vowels

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number of words $-$ number of words with exactly zero vowels = $(\\var{v} + \\var{c})^\\var{w} - \\var{c}^\\var{w} = \\var{(v+c)^w - answers[0]}$.

", "statement": "

Suppose you have a hypothetical new alphabet with $\\var{c+v}$ letters: $\\var{c}$ consonants and $\\var{v}$ vowels.

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Note that the NUMBAS syntax for $C(n,r)$ is comb(n,r).

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How many different $\\var{w}$-letter words can you make with this alphabet?

Fill out the following table

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
 $n$ Number of $\\var{w}$-letter words with exactly $n$ vowels $0$ [[0]] $1$ [[1]] $2$ [[2]] $3$ [[3]] $4$ [[4]] $5$ [[5]] TOTAL [[6]]
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It may be simpler to take the total number of words, and subtract the number of words with no vowels.

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How many $\\var{w}$-letter words have at least one vowel?

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