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Jlenok [28]
2 years ago
7

Pat had 9 flowerpots, and she wants to plant a different type of flower in each one. There are 11 types of flowers available at

the garden shop. In how many different ways can she choose the flower
Mathematics
2 answers:
wariber [46]2 years ago
7 0

<span>In this problem, we will use the combination since the order does not matter but the flower can only be selected once. From the given, we have a total of 11 flower and 9 flowerpots, to get the number of possible combinations, we can write is at 11C9 or 11! / {9! (11-9)!}. The total number of possible combinations is 55</span>

matrenka [14]2 years ago
3 0

Answer:

The answer is 55

Step-by-step explanation:

The order doesn't matter so its just 55.

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Two armadillos and three aardvarks sat in a five-seat row at the Animal Auditorium. What is the probability that the two animals
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Answer:

      \dfrac{1}{15}

Explanation:

The number of different ways in which the<em> two armadillos</em> would be at the ends of the row is 2:

                      P(2,2)=\dfrac{2!}{(2-2)!}=\dfrac{2}{0!}=2

The number of different combinations in which<em> two of the three aardvarks </em>can sit at the ends of the row is P(3,2):

          P(3,2)=\dfrac{3!}{(3-2)!}=\dfrac{3\times 2\times 1}{1}=6

Therefore, there are 2 + 6 = 8 different ways in which the two animlas on the ends of the row were both armadillos or both aardvarks.

Now calculate the total number of different ways in which the animals can sit. It is P(5,5):

        P(5,5)=\dfrac{5!}{(5-5)!}=5!=5\times 4\times 3\times 2\times 1=120

Thus, <em>the probability that the two animals on the ends of the row were both armadillos or both aardvarks</em> is equal to the number of favorable outputs divided by the total number of possible outputs:

      Probability=\dfrac{8}{120}=\dfrac{1}{15}

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