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Serga [27]
2 years ago
14

How many three-letter "words" can be made from 5 letters "FGHIJ" if repetition of letters

Mathematics
1 answer:
pashok25 [27]2 years ago
5 0

The number of ways of the letters FGHIJ can be arranged with repetition and without repetition are 125 and 60 ways respectively.

<h3>How to determine the ways </h3>

Formula for permutation with repetition

^nPr = n^r

Formula for permutation without repetition

^nPr = \frac{n!}{n-r!}

We have = 5

r = 3

a. If repetition is allowed,

Permutation = 5^3 = 125 ways

b. If repetition is not allowed,

Permutation = \frac{5!}{5-3!}

Permutation = \frac{5*4*3*2*1}{2*1}

Permutation = \frac{120}{2}

Permutation = 60 ways

Thus, the number of ways of the letters FGHIJ can be arranged with repetition and without repetition are 125 and 60 ways respectively.

Learn more about permutation here:

brainly.com/question/12468032

#SPJ1

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=?

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search-icon-image

Class 9

>>Maths

>>Quadrilaterals

>>Quadrilaterals and Their Various Types

>>In Fig. 6.43, if PQ PS, PQ∥ SR, SQR = 2

Question

Bookmark

In Fig. 6.43, if PQ⊥PS,PQ∥SR,∠SQR=28

0

and ∠QRT=65

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, then find the values of x and y.

463685

expand

Medium

Solution

verified

Verified by Toppr

Given, PQ⊥PS,PQ∥SR,∠SQR=28

∘

,∠QRT=65

∘

According to the question,

x+∠SQR=∠QRT (Alternate angles as QR is transversal.)

⇒x+28

∘

=65

∘

⇒x=37

∘

Also ∠QSR=x

⇒∠QSR=37

∘

Also ∠QRS+∠QRT=180

∘

(Linear pair)

⇒∠QRS+65

∘

=180

∘

⇒∠QRS=115

∘

Now, ∠P+∠Q+∠R+∠S=360

∘

(Sum of the angles in a quadrilateral.)

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+65

∘

+115

∘

+∠S=360

∘

⇒270

∘

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∘

⇒270

∘

+y+37

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⇒307

∘

+y=360

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⇒y=53

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