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Ritu K. Class I-V Tuition trainer in Mumbai/>

Ritu K.

Maths Tuitor

Kanjurmarg West, Mumbai, India - 400078.

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Overview

Ritu K. describes herself as Maths Tuitor. She conducts classes in Class 11 Tuition, Class 12 Tuition and Class I-V Tuition. Ritu is located in Kanjurmarg West, Mumbai. Ritu takes at students Home. She is well versed in Hindi and English.

Languages Spoken

Hindi

English

Address

Kanjurmarg West, Mumbai, India - 400078

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Teaches

Class I-V Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

CBSE, State

CBSE Subjects taught

Mathematics

Taught in School or College

No

Class 11 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

CBSE

CBSE Subjects taught

Mathematics

Taught in School or College

No

Class 12 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

CBSE

CBSE Subjects taught

Mathematics

Taught in School or College

No

Reviews

No Reviews yet!

FAQs

1. Which school boards of Class 1-5 do you teach for?

CBSE and State

2. Have you ever taught in any School or College?

No

3. Which classes do you teach?

I teach Class 11 Tuition, Class 12 Tuition, Class I-V Tuition and Class VI-VIII Tuition Classes.

4. Do you provide a demo class?

Yes, I provide a free demo class.

5. How many years of experience do you have?

I have been teaching for less than a year.

Contact

Answers by Ritu K. (4)

Answered on 24/10/2016 Learn Tuition/Class XI-XII Tuition (PUC) +2 CBSE/Class 12/Mathematics Permutations

Using all the letters of the word "FRIDAY", how many different words can be formed?

As the given word "FRIDAY" in the question has 6 letters, so the total number of words that can be formed is a factorial of 6. 6! = 720
Answers 50 Comments
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Answered on 24/10/2016 Learn Tuition/Class XI-XII Tuition (PUC) +2 CBSE/Class 12/Mathematics Permutations

What is the value of 6P3?

6P3 = 6!/(6-3)! =120
Answers 27 Comments
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Answered on 24/10/2016 Learn Tuition/Class XI-XII Tuition (PUC) +2 CBSE/Class 12/Mathematics Permutations

The number of sequences in which 9 players can throw a volley ball, so that the youngest player may not... ...more
The number of sequences in which 9 players can throw a volley ball, so that the youngest player may not be the last is ____ ?

Total number of ways to sequence 9 players = 9! Total number of ways to sequence 9 players with the youngest boy at the last position = 8! (It is 8! because the position of the youngest is fixed i.e. at the last place) So, the Number of ways that the youngest player will not be at the last is = 9... ...more
Total number of ways to sequence 9 players = 9! Total number of ways to sequence 9 players with the youngest boy at the last position = 8! (It is 8! because the position of the youngest is fixed i.e. at the last place) So, the Number of ways that the youngest player will not be at the last is = 9!-8!=322560
Answers 14 Comments
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Answered on 24/10/2016 Learn Tuition/Class XI-XII Tuition (PUC) +2 CBSE/Class 12/Mathematics Permutations

Out of 12 consonants and 4 vowels, how many words of 6 consonants and 3 vowels can be formed?

Number of ways to select 6 consonants from 12 = 12C6 Number of ways to select 3 vowels from 4 = 4C3 Number of ways of selecting 6 consonants from 12 and 3 vowels from 4 = 12C6 × 4C3 =(12!/((12-6)!*6!)) ×(4!/((4-3)!*4!)) =(12!/(6!*6!) ×(4!/(1!*4!)) =3696 It means we can have 3696 groups where... ...more
Number of ways to select 6 consonants from 12 = 12C6 Number of ways to select 3 vowels from 4 = 4C3 Number of ways of selecting 6 consonants from 12 and 3 vowels from 4 = 12C6 × 4C3 =(12!/((12-6)!*6!)) ×(4!/((4-3)!*4!)) =(12!/(6!*6!) ×(4!/(1!*4!)) =3696 It means we can have 3696 groups where each group contains total 9 letters (6 consonants and 3 vowels). Number of ways to arrange 9 letters among themselves =9! Hence, required number of ways =3696×9!
Answers 14 Comments
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Contact

Teaches

Class I-V Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

CBSE, State

CBSE Subjects taught

Mathematics

Taught in School or College

No

Class 11 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

CBSE

CBSE Subjects taught

Mathematics

Taught in School or College

No

Class 12 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

CBSE

CBSE Subjects taught

Mathematics

Taught in School or College

No

No Reviews yet!

Answers by Ritu K. (4)

Answered on 24/10/2016 Learn Tuition/Class XI-XII Tuition (PUC) +2 CBSE/Class 12/Mathematics Permutations

Using all the letters of the word "FRIDAY", how many different words can be formed?

As the given word "FRIDAY" in the question has 6 letters, so the total number of words that can be formed is a factorial of 6. 6! = 720
Answers 50 Comments
Dislike Bookmark

Answered on 24/10/2016 Learn Tuition/Class XI-XII Tuition (PUC) +2 CBSE/Class 12/Mathematics Permutations

What is the value of 6P3?

6P3 = 6!/(6-3)! =120
Answers 27 Comments
Dislike Bookmark

Answered on 24/10/2016 Learn Tuition/Class XI-XII Tuition (PUC) +2 CBSE/Class 12/Mathematics Permutations

The number of sequences in which 9 players can throw a volley ball, so that the youngest player may not... ...more
The number of sequences in which 9 players can throw a volley ball, so that the youngest player may not be the last is ____ ?

Total number of ways to sequence 9 players = 9! Total number of ways to sequence 9 players with the youngest boy at the last position = 8! (It is 8! because the position of the youngest is fixed i.e. at the last place) So, the Number of ways that the youngest player will not be at the last is = 9... ...more
Total number of ways to sequence 9 players = 9! Total number of ways to sequence 9 players with the youngest boy at the last position = 8! (It is 8! because the position of the youngest is fixed i.e. at the last place) So, the Number of ways that the youngest player will not be at the last is = 9!-8!=322560
Answers 14 Comments
Dislike Bookmark

Answered on 24/10/2016 Learn Tuition/Class XI-XII Tuition (PUC) +2 CBSE/Class 12/Mathematics Permutations

Out of 12 consonants and 4 vowels, how many words of 6 consonants and 3 vowels can be formed?

Number of ways to select 6 consonants from 12 = 12C6 Number of ways to select 3 vowels from 4 = 4C3 Number of ways of selecting 6 consonants from 12 and 3 vowels from 4 = 12C6 × 4C3 =(12!/((12-6)!*6!)) ×(4!/((4-3)!*4!)) =(12!/(6!*6!) ×(4!/(1!*4!)) =3696 It means we can have 3696 groups where... ...more
Number of ways to select 6 consonants from 12 = 12C6 Number of ways to select 3 vowels from 4 = 4C3 Number of ways of selecting 6 consonants from 12 and 3 vowels from 4 = 12C6 × 4C3 =(12!/((12-6)!*6!)) ×(4!/((4-3)!*4!)) =(12!/(6!*6!) ×(4!/(1!*4!)) =3696 It means we can have 3696 groups where each group contains total 9 letters (6 consonants and 3 vowels). Number of ways to arrange 9 letters among themselves =9! Hence, required number of ways =3696×9!
Answers 14 Comments
Dislike Bookmark

Contact

Ritu K. describes herself as Maths Tuitor. She conducts classes in Class 11 Tuition, Class 12 Tuition and Class I-V Tuition. Ritu is located in Kanjurmarg West, Mumbai. Ritu takes at students Home. She is well versed in Hindi and English.

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