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2 years ago

Find the value of 13! / 3!10!

Mathematics

2 answers:

snow_lady [41]2 years ago

5 0

Answer:

The answer is 286.

Step-by-step explanation:

13! can be rewritten as 13*12*11*10!.

Since 10! is on the top and bottom of the fraction, we can eliminate them. We are then left with 13*12*11/3!.

3! = 6. We can eliminate the 6 from the denominator by diving the 12 by 6 in the numerator to get 13*11*2.

Finally, we can can evaluate the expression to get 286.

djyliett [7]2 years ago

5 0

Answer:

the answer is 286

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MaRussiya [10]

Answer: none of the above

Step-by-step explanation: when performing an hypothesis test and we want to make conclusion by comparing the p-value with the level of significance α

When p is greater than α, we reject the null hypothesis because it simply implies that we have a larger chance to commit a type 1 error ( α is the probability of committing a type 1 error an error where we reject the null hypothesis instead of accepting it ) which means we reject the null hypothesis.

When p is lesser than level of significance α, it means that we have a lesser chance of committing a type 1 error, which means we accept the null hypothesis.

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3 years ago

Evaluate the expression. StartFraction 9 factorial Over 3 factorial EndFraction 3 6 60,480 362,874

REY [17]

Answer:

60,480 is the correct answer.

Step-by-step explanation:

First of all, let us have a look at the formula of factorial of a number 'n':

$n! = n \times (n-1) \times (n-2) \times ...... \times 1$

i.e. multiply n with (n-1) then by (n-2) upto 1.

Keep on subtracting 1 from the number and keep on multiplying until we reach to 1.

So, $9!$ can be written as: $9 \times 8 \times 7 \times ...... \times 1$

Similarly $3!$ can be written as: $3 \times 2 \times 1$

Re-writing $9 !$ :

$9 \times 8 \times 7 \times ...... 3 \times 2 \times 1\\\Rightarrow 9 \times 8 \times 7 \times ...... 3 !$

Now, the expression to be evaluated:

$\dfrac{9!}{3!} = \dfrac{9 \times 8 \times 7 \times ..... \times 3!}{3!}\\\Rightarrow 9 \times 8 \times 7 \times 6 \times 5 \times 4\\\Rightarrow 60480$

5 0

2 years ago

Read 2 more answers

What value of x makes this equation true?-80=20+x

Svet_ta [14]

Answer:

50 i think

Step-by-step explanation:

4 0

2 years ago

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Eight people attend the theater "together and sit in a row with exactly eight seats. Suppose the eight people consist of three m

Nata [24]

Answer:

They can be seated in 120 differents ways.

Step-by-step explanation:

Taking into account that there are 3 couples and every couple has an specific way to sit, for simplify the exercise, every couple is going to act like 1 option and it's going to occupy 1 Place. If this happens we just need to organize 5 options (3 couples and 2 singles) in 5 Places (3 for a couple and 2 for the singles)

It means that now there are just 5 Places in the row and 5 options to organized. So the number of ways can be calculated using a rule of multiplication as:

5 * 4 * 3 * 2 * 1 = 120

1st place 2nd Place 3rd place 4th Place 5th Place

Because we have 5 options for the 1st Place, the three couples and the 2 singles. Then, 4 options for the second Place, 3 options for the third place, 2 for the fourth place and 1 option for the 5th place.

Finally, they can be seated in 120 differents ways.

8 0

2 years ago

A donut store has 11 different types of donuts. You can only buy a bag of 3 of them, where each donut has to be of a different t

MakcuM [25]

Answer:

$165$ .

Step-by-step explanation:

Since repetition isn't allowed, there would be $11$ choices for the first donut, $(11 - 1) = 10$ choices for the second donut, and $(11 - 2) = 9$ choices for the third donut. If the order in which donuts are placed in the bag matters, there would be $11 \times 10 \times 9$ unique ways to choose a bag of these donuts.

In practice, donuts in the bag are mixed, and the ordering of donuts doesn't matter. The same way of counting would then count every possible mix of three donuts type $3 \times 2 \times 1 = 6$ times.

For example, if a bag includes donut of type $x$ , $y$ , and $z$ , the count $11 \times 10 \times 9$ would include the following $3 \times 2 \times 1$ arrangements:

$xyz$ .
$xzy$ .
$yxz$ .
$yzx$ .
$zxy$ .
$zyx$ .

Thus, when the order of donuts in the bag doesn't matter, it would be necessary to divide the count $11 \times 10 \times 9$ by $3 \times 2 \times 1 = 6$ to find the actual number of donut combinations:

$\begin{aligned} \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}$ .

Using combinatorics notations, the answer to this question is the same as the number of ways to choose an unordered set of $3$ objects from a set of $11$ distinct objects:

$\begin{aligned}\begin{pmatrix}11 \\ 3\end{pmatrix} &= \frac{11 !}{(11 - 3)! \times 3 !} \\ &= \frac{11 !}{8 ! \times 3 !} \\ &= \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}$ .

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1 year ago