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Sagot :
In this problem we are given that a person who wants to go to a party would like to bring 3 different bags of chips; and also, that he/she has 13 varieties.
As she has differents options to choose, we see that this one is a problem of permutations or combinations. But the order of the chips the person brings does not matter, and so, we have to find a combination. We remember then that the equation for finding a combination of n objects and k possibilites is given by the following binomial coefficient
[tex]\binom{n}{k}=\frac{n!}{k!(n-k)!}[/tex]Where the symbol ! represents factorial (the product of the number to the number). For example,
[tex]4!=4\cdot3\cdot2\cdot1=24[/tex]Now, in our exercise, we have that the person has 13 varieties and has to choose 3 of them. We have to calculate then:
[tex]\binom{13}{3}[/tex]For doing so, we will start by replacing the values on the formula:
[tex]\begin{gathered} \binom{13}{3}=\frac{13!}{3!(13-3)!} \\ =\frac{13!}{3!(10)!} \\ =\frac{13\cdot12\cdot11\cdot10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{3\cdot2\cdot10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1} \end{gathered}[/tex]And cancelling the equal terms, we obtain:
[tex]\begin{gathered} =\frac{13\cdot12\cdot11\cdot\cancel{10}\cdot\cancel{9}\cdot\cancel{8}\cdot\cancel{7}\cdot\cancel{6}\cdot\cancel{5}\cdot\cancel{4}\cdot\cancel{3}\cdot\cancel{2}\cdot\cancel{1}}{3\cdot2\cdot\cancel{10}\cdot\cancel{9}\cdot\cancel{8}\cdot\cancel{7}\cdot\cancel{6}\cdot\cancel{5}\cdot\cancel{4}\cdot\cancel{3}\cdot\cancel{2}\cdot\cancel{1}} \\ =\frac{13\cdot12\cdot11}{3\cdot2} \\ =\frac{13\cdot12\cdot11}{6} \\ =13\cdot2\cdot11=286 \end{gathered}[/tex]This means that the person would be able to do 286 different selections.
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