All categories

3 years ago

How many distinguishable ways can you arrange the letters in the word "MISSISSIPPI"?

Mathematics

2 answers:

Flauer [41]3 years ago

7 0

Answer: 34,650

Step-by-step explanation:

Given Word : "MISSISSIPPI"

The total number of letters in the given word: 11

Number of times S appears : 4

Number of times I appears : 4

Number of times P appears :2

Now, by permutations the number of arrangements of the letters in the given word will be :

$n=\dfrac{11!}{4!4!2!}\\\\=34650$

Hence, the the number of arrangements of the letters in the given word= 34,650.

Ipatiy [6.2K]3 years ago

3 0

34,650 is the answer, I think

You might be interested in

Lines DE and AB intersect at point C.

alexandr402 [8]

Answer:

Step-by-step explanation:

2x+2+5x+3=180 (angles on a straight line)

7x+5=180

7x=180-5

7x=175

x=25 (dividing both sides by 7)

3 0

2 years ago

Find the Arc Length and Sector Area

gulaghasi [49]

Answer:

Need points

Step-by-step explanation:

Need points

5 0

2 years ago

Solve the equation 5x-2x^2+1

Pani-rosa [81]

Answer:

Step-by-step explanation:

If you call "5x-2x^2+1" an "equation," then you must equate 5x-2x^2+1 to 0:

5x-2x^2+1 = 0

This is a quadratic equation. Rearranging the terms in descending order by powers of x, we get:

-2x^2 + 5x + 1 = 0. Here the coefficients are a = -2, b = 5 and c = 1.

Use the quadratic formula to solve for x:

First find the discriminant, b^2 - 4ac: 25 - 4(-2)(1) = 25 + 8 = 33

Because the discriminant is positive, the roots of this quadratic are real and unequal.

-b ± √(discriminant)

Applying the quadratic formula x = --------------------------------

we get:

-5 ± √33 -5 + √33

x = ----------------- = --------------------- and

2(-2) -4

-5 - √33

---------------

-4

5 0

2 years ago

Question : In the given figure , ∆ APB and ∆ AQC are equilateral triangles. Prove that PC = BQ.

lorasvet [3.4K]

Answer:

See Below.

Step-by-step explanation:

We are given that ΔAPB and ΔAQC are equilateral triangles.

And we want to prove that PC = BQ.

Since ΔAPB and ΔAQC are equilateral triangles, this means that:

$PA\cong AB\cong BP\text{ and } QA\cong AC\cong CQ$

Likewise:

$\angle P\cong \angle PAB\cong \angle ABP\cong Q\cong \angle QAC\cong\angle ACQ$

Since they all measure 60°.

Note that ∠PAC is the addition of the angles ∠PAB and ∠BAC. So:

$m\angle PAC=m\angle PAB+m\angle BAC$

Likewise:

$m\angle QAB=m\angle QAC+m\angle BAC$

Since ∠QAC ≅ ∠PAB:

$m\angle PAC=m\angle QAC+m\angle BAC$

And by substitution:

$m\angle PAC=m\angle QAB$

Thus:

$\angle PAC\cong \angle QAB$

Then by SAS Congruence:

$\Delta PAC\cong \Delta BAQ$

And by CPCTC:

$PC\cong BQ$

5 0

2 years ago

Read 2 more answers

1/2n-17/12=2/3n-13/6

emmasim [6.3K]

1/2/-33+2=62 but I d k

8 0

3 years ago

Read 2 more answers