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Sanjay Krishna Class 11 Tuition trainer in Bangalore

Sanjay Krishna

locationImg Teachers Colony, HSR Layout, Bangalore
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Mathematics and Physics tutor

Student's home
I have tutored students in class 8th and 9th to find out that there is a huge gap between what is taught and what is learn t. Learning enhances the way that you look at the world. Looks into the details we don't understand otherwise. I wish to bring that awareness to a student's mind. My strengths lie in the ability to breakdown large problems into smaller solvable blocks, which i intend to teach. That will enable students to just not see the solution but be able to solve tougher problems by themselves.

Languages Spoken

Malayalam

English

Hindi

Education

Amity University, Noida 2009

Bachelor of Technology (B.Tech.)

Address

Teachers Colony, HSR Layout, Bangalore, India - 560034

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Teaches

Class 11 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

ISC/ICSE, IGCSE, CBSE

Subjects taught

Computer Science, Mathematics, Physics

Taught in School or College

No

Class 12 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

ISC/ICSE, IGCSE, CBSE

Subjects taught

Computer Science, Physics, Mathematics

Taught in School or College

No

Class 9 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

CBSE, IGCSE, ICSE

Subjects taught

Information Technology, Computer Application, Mathematics, Physics, Computer Practices, Science

Taught in School or College

No

Class 10 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

CBSE, IGCSE, ICSE

Subjects taught

Computer Application, Computer Practices, Physics, Mathematics, Information Technology, Science

Taught in School or College

No

Reviews

No Reviews yet!

Answers by Sanjay Krishna

Answered on 05/10/2016 Learn Tuition/Class XI-XII Tuition (PUC) +2 CBSE - Class 12/Mathematics Permutations

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Post a Lesson

Key 1 : Any number whose digits add up to a multiple of 3 is divisible by 3. 6 digit combinations divisible by 3 : A. {0,1,2,3,4,5} => 0+1+2+3+4+5 = 15 (divisible by 3) total number of 6 digit integers formed with the above combination=5*5*4*3*2*1 = 600 B. {1,2,3,4,5,6} => 1+2+3+4+5+6 =... ...more
Key 1 : Any number whose digits add up to a multiple of 3 is divisible by 3. 6 digit combinations divisible by 3 : A. {0,1,2,3,4,5} => 0+1+2+3+4+5 = 15 (divisible by 3) total number of 6 digit integers formed with the above combination=5*5*4*3*2*1 = 600 B. {1,2,3,4,5,6} => 1+2+3+4+5+6 = 21 (divisible by 3) total number of 6 digit integers formed with the above combination=6*5*4*3*2*1 = 720 C. {0,1,2,4,5,6} => 0+1+2+4+5+6 = 18 (divisible by 3) total number of 6 digit integers formed with the above combination=5*5*4*3*2*1 = 600 From A, B and C total number of integers=> 600+720+600 = 1920
Answers 24 Comments
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Teaches

Class 11 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

ISC/ICSE, IGCSE, CBSE

Subjects taught

Computer Science, Mathematics, Physics

Taught in School or College

No

Class 12 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

ISC/ICSE, IGCSE, CBSE

Subjects taught

Computer Science, Physics, Mathematics

Taught in School or College

No

Class 9 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

CBSE, IGCSE, ICSE

Subjects taught

Information Technology, Computer Application, Mathematics, Physics, Computer Practices, Science

Taught in School or College

No

Class 10 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

CBSE, IGCSE, ICSE

Subjects taught

Computer Application, Computer Practices, Physics, Mathematics, Information Technology, Science

Taught in School or College

No

No Reviews yet!

Answers by Sanjay Krishna

Answered on 05/10/2016 Learn Tuition/Class XI-XII Tuition (PUC) +2 CBSE - Class 12/Mathematics Permutations

Ask a Question

Post a Lesson

Key 1 : Any number whose digits add up to a multiple of 3 is divisible by 3. 6 digit combinations divisible by 3 : A. {0,1,2,3,4,5} => 0+1+2+3+4+5 = 15 (divisible by 3) total number of 6 digit integers formed with the above combination=5*5*4*3*2*1 = 600 B. {1,2,3,4,5,6} => 1+2+3+4+5+6 =... ...more
Key 1 : Any number whose digits add up to a multiple of 3 is divisible by 3. 6 digit combinations divisible by 3 : A. {0,1,2,3,4,5} => 0+1+2+3+4+5 = 15 (divisible by 3) total number of 6 digit integers formed with the above combination=5*5*4*3*2*1 = 600 B. {1,2,3,4,5,6} => 1+2+3+4+5+6 = 21 (divisible by 3) total number of 6 digit integers formed with the above combination=6*5*4*3*2*1 = 720 C. {0,1,2,4,5,6} => 0+1+2+4+5+6 = 18 (divisible by 3) total number of 6 digit integers formed with the above combination=5*5*4*3*2*1 = 600 From A, B and C total number of integers=> 600+720+600 = 1920
Answers 24 Comments
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