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Assume that the student has a cup with 12 writing implements: 6 pencils, 4 ball point pens, and 2 felt-tip pens. In how many ways can the selection be made if no more than one ball point pen is selected if they must choose 4 implements?

Guest Sep 5, 2017
 #1
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cool cool cool

CPhill  Sep 5, 2017
edited by CPhill  Sep 5, 2017
edited by CPhill  Sep 6, 2017
 #2
avatar+19482 
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Assume that the student has a cup with 12 writing implements: 6 pencils, 4 ball point pens, and 2 felt-tip pens. In how many ways can the selection be made if no more than one ball point pen is selected if they must choose 4 implements?

 

\(\begin{array}{|rcll|} \hline && \binom{4}{1}\binom{8}{3} + \binom{4}{0}\binom{8}{4} \\ &=& 4\binom{8}{3} + 1\binom{8}{4} \\ &=& 4\cdot \frac{8}{3}\cdot \frac{7}{2}\cdot \frac{6}{1} + \frac{8}{4}\cdot \frac{7}{3}\cdot \frac{6}{2}\cdot \frac{5}{1} \\ &=& 4\cdot 8 \cdot 7 + 7 \cdot 2 \cdot 5 \\ &=& 224 + 70 \\ &=& 294 \\ \hline \end{array}\)

 

laugh

heureka  Sep 6, 2017

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