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lubasha [3.4K]
2 years ago
6

comparing permutations to combinations for the same set of parameters you would have more combinations than permutations true or

false
Mathematics
2 answers:
ella [17]2 years ago
6 0
Let a set of n elements. 
We can find n! (factorial) of the n element. 
However, combination of the element lead to less than n! possibilities. 
(combining like adding or multiplying)
So the proposition is false. 
Illusion [34]2 years ago
3 0

Answer:

False; you would have more permutations than combinations.

Step-by-step explanation:

The formula for taking combinations of <em>n</em> objects taken <em>r </em>at a time is

\frac{n!}{r!(n-r)!}

The formula for taking permutations of <em>n</em> objects taken <em>r</em> at a time is

\frac{n!}{(n-r)!}

Comparing these two, we can see that the difference between the formulas is that the formula for combinations is divided by an extra r!.  Since it is divided by a larger number, it will result in a smaller answer; therefore permutations give more results than combinations.

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Tomorrow, Mrs. Wendel's class will be using toothpicks for a science project. Each student must use at least 9 toothpicks for th
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Number of toothpicks required per student = At least 9

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Hitung jarak di antara titik P(7, 1) dengan titik Q(7, 8) ​
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7 units

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=\sqrt{(y1 - y2)^{2}  + (x1 - x2) {}^{2} }

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5 0
1 year ago
3. A CD costs £9.50 in London and
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Only Q4 has clear details

therefore, solving Question 4,

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6 0
2 years ago
) for a particular diamond mine, 77% of the diamonds fail to qualify as "gemstone grade". a random sample of 112 diamonds is ana
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z = (x – u) / s

 

But first let us calculate the standard deviation s, sample x and mean u.

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s = sqrt (112 * 0.77 * (1 – 0.77))

s = 4.45

 

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u = 0.77 * 112 = 86.24

 

So the z score is:

z = (90.72 – 86.24) / 4.45

z = 1.00

 

From the standard tables, the P value at z = 1.00 using right tailed test is:

<span>P = 0.1587 = 15.87%</span>

7 0
1 year ago
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