Answered on 11/12/2016 CBSE/Class 12/Science/Mathematics Tuition/Class XI-XII Tuition (PUC) Permutations

K.vaishali .

4x5x6x7x8=6720.

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Answers 20 Comments Answered on 28/11/2016 CBSE/Class 12/Science/Mathematics Tuition/Class XI-XII Tuition (PUC) Permutations

Debojyoti Roy

Tutor

118.

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Answers 6 Comments Answered on 06/12/2016 CBSE/Class 12/Science/Mathematics Tuition/Class XI-XII Tuition (PUC) Permutations

Sarvajeet Kumar

An Experienced Trainer

8.

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Answers 3 Comments Answered on 06/12/2016 CBSE/Class 12/Science/Mathematics Tuition/Class XI-XII Tuition (PUC) Permutations

Sarvajeet Kumar

An Experienced Trainer

336.

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Answers 3 Comments Answered on 02/12/2016 CBSE/Class 12/Science/Mathematics Tuition/Class XI-XII Tuition (PUC) Permutations

Abhilash

Teacher

In mathematics, the notion of permutation relates to the act of arranging all the members of a set into some sequence or order, or if the set is already ordered, rearranging (reordering) its elements, a process called permuting.

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Answers 3 Comments Answered on 06/12/2016 CBSE/Class 12/Science/Mathematics Tuition/Class XI-XII Tuition (PUC) Permutations

Rohit Singh

Commerce Master

9*8*7=504 types number can formed by using 1 to 9 digit of 3 digit numbers without repeated number.

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Answers 4 Comments

Sarvajeet Kumar

An Experienced Trainer

6! = 720.

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Answers 3 Comments Answered on 30/11/2016 CBSE/Class 12/Science/Mathematics Tuition/Class XI-XII Tuition (PUC) Permutations

Given 5 flags of different colours, how many different signals can be generated if each signal requires... read more

Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other? read less

Abhilash Sharma

Teacher for 8th-12th)

5C2 *2! = 20 Out of 5 flags choosing any two will give you a signal. Now these flags can be interchanged (suppose red on top and blue below it, now blue on top and red below it) Hence 2! for this arrangement among these two colours themselves.

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Answers 2 Comments Answered on 12/10/2016 CBSE/Class 12/Science/Mathematics Permutations Tuition/Class XI-XII Tuition (PUC)

Durgesh Kumar

Tutor

Q;- FRIDAY Total no of different letters=F+R+I+D+A+Y = 6! = 6*5*4*3*2*1 = 720

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Answers 47 Comments Answered on 23/10/2016 CBSE/Class 12/Science/Mathematics Permutations Tuition/Class XI-XII Tuition (PUC)

Debojyoti Roy

Tutor

7!/(7-6)! =7!/1! =7*6*5*4*3*2*1

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