All categories

2 years ago

How many distinct permutations can be made from the letters of the word independence?.

SAT

1 answer:

Komok [63]2 years ago

3 0

There are 1,663,200 different permutations.

<h3>How many different permutations can be made?</h3>

First, we need to count the number of different elements in the word independence.

We have 6 different letters.

i appears only one time.
n appears 3 times.
e appears 4 times.
d appears 2 times.
p appears one time.
c appears one time.

For a total of 12 letters. The different number of permutations is given by the factorial of the total number of elements (12) divided by the factorials of the number of repetitions of all the elements. Then we have:

$P = \frac{12!}{1!*3!*4!*2!*1!*1!} = \frac{12!}{3!*4!*2!} = \frac{12*11*10*9*8*7*6*5}{3*2*2} = 1,663,200$

So there are 1,663,200 different permutations.

If you want to learn more about permutations, you can read:

brainly.com/question/12468032

You might be interested in

Which of the following phrases best describes ethnocentrism

Alex73 [517]

Answer:

what are the phrases??????????

4 0

3 years ago

Consider a set of cards that has four cards labeled 1, 2, 3, and 4. Suppose you pick two cards, without replacement, to obtain t

Phantasy [73]

Answer:

$\begin{array}{cccccc}{Mean} & {1.5} & {2} & {2.5} & {3} & {3.5}\ \\ {Probability} & {\frac{1}{6}} & {\frac{1}{6}} & {\frac{1}{3}} & {\frac{1}{6}} & {\frac{1}{6}}\ \end{array}$

Explanation:

Given

$Cards = \{1,2,3,4\}$

Required

The sampling distribution

The possible selection of 2 cards without replacement is as follows:

$S = \{(1,2) (1,3) (1,4) (2,1) (2,3) (2,4) (3,1) (3,2) (3,4) (4,1) (4,2) (4,3)\}$

Calculate the mean

$\begin{array}{cccccccccccc}{Selection} & {(1,2)} & {(1,3)} & {(1,4)} & {(2,1)} & {(2,3)} & {(2,4)}& {(3,1)} & {(3,2)} & {(3,4)} & {(4,1)} & {(4,2)} & {(4,3)} \ \\ {Mean} & {1.5} & {2} & {2.5} & {1.5} & {2.5} & {3} & {2} & {2.5} & {3.5} & {2.5} & {3} & {3.5}\ \end{array}$

List out the mean and the respective frequency

$1.5 \to 2$

$2 \to 2$

$2.5 \to 4$

$3 \to 2$

$3.5\to 2$

$Total \to 12$

Calculate the probability of each mean

$P(1.5) \to \frac{2}{12} \to \frac{1}{6}\\$

$P(2) \to \frac{2}{12} \to \frac{1}{6}\\$

$P(2.5) \to \frac{4}{12} \to \frac{1}{3}\\$

$P(3) \to \frac{2}{12} \to \frac{1}{6}\\$

$P(3.5) \to \frac{2}{12} \to \frac{1}{6}$

So, the table of sampling distribution is:

$\begin{array}{cccccc}{Mean} & {1.5} & {2} & {2.5} & {3} & {3.5}\ \\ {Probability} & {\frac{1}{6}} & {\frac{1}{6}} & {\frac{1}{3}} & {\frac{1}{6}} & {\frac{1}{6}}\ \end{array}$

3 0

3 years ago

A patient who has had multiple blood draws from the same vein may notice some inflammation in the area. This is referred to as:

amid [387]

A patient who has had multiple bloods draws from the same vein may notice some inflammation in the area. This is referred to as phlebitis. This is further explained below.

<h3>What is phlebitis?</h3>

Generally, vein wall inflammation also known as varicose veins.

In conclusion, A patient who has had blood drawn from the same vein several times may feel some irritation. Phlebitis is the medical term for this.