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Gelneren [198K]
3 years ago
9

COMBINATION/PERMUTATION MATH HELP???

Mathematics
1 answer:
Artyom0805 [142]3 years ago
5 0
This type of combination problem involves combined probability, or the chance that a specific set could be chosen based on the probability of multiple variables.

The number that could be generated follows the example:

XXYZZZZ

where X describes a letter, and Y describes a number that isn’t zero, and Z describes any number.

Two probabilities exist in this situation, depending on the circumstances of the question. By pressing any number once, only 8 letters can be used (ten digits, minus the one and two keys, since they don’t have letters). This means that the probability of this event is:

8 x 8 x 9 x 10 x 10 x 10 x 10

where the two eights are for the letters, the nine is for the digit that isn’t zero, and the tens are for any numbers from the keypad. This permutation yields 5,760,000 choices.

If you are allowed to press the numbers more than once to generate a letter, then the probability changes to account for the entire alphabet. The new probability of this event is :

26 x 26 x 9 x 10 x 10 x 10 x 10

where this permutation yields 60,840,000 choices.

Hope this helps!
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To find which expression is equal to the given expression.

$\frac{\left(5 g^{4}+5 g^{3}-17 g^{2}+6 g\right)-\left(3 g^{4}+6 g^{3}-7 g^{2}-12\right)}{g+2}

Expand the term -\left(3 g^{4}+6 g^{3}-7 g^{2}-12\right):-3 g^{4}-6 g^{3}+7 g^{2}+12

         $=\frac{5 g^{4}+5 g^{3}-17 g^{2}+6 g- 3 g^{4}-6 g^{3}+7 g^{2}+12}{g+2}

Arrange the like terms together.

         $=\frac{5 g^{4}- 3 g^{4}+5 g^{3}-6 g^{3}-17 g^{2}+7 g^{2}+6 g+12}{g+2}

         $=\frac{2 g^{4}- g^{3}-10 g^{2}+6 g+12}{g+2}

Factor the numerator 2 g^{4}-g^{3}-10 g^{2}+6 g+12=(g+2)\left(2 g^{3}-5 g^{2}+6\right)

         $=\frac{(g+2)\left(2 g^{3}-5 g^{2}+6\right)}{g+2}

Cancel the common factor g + 2, we get

         =2 g^{3}-5 g^{2}+6

Hence option C is the correct answer.

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