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Sarvajeet

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D

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P

Prabha replied | 11/12/2016

8!-3!=8×7×6×5×4×3!- 3!= 3!(8×7×6×5×4-1)
=6(6720-1)=6×6719=40314.

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K

K.vaishali replied | 11/12/2016

4x5x6x7x8=6720.

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A

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Daljit Kaur replied | 24/11/2016

64.

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Hassan replied | 05/12/2016

1/6! +1/(7x6!) =1/(8x7x6!). Take LCM and solve will get x=64.

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Debojyoti Roy replied | 28/11/2016

118.

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Sarvajeet replied | 06/12/2016

118.

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B

Bhuvana 10/11/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC)), Permutations

Evaluate: 8! / 6! * 2!?

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Tapas replied | 01/12/2016

The problem should be mentioned more clearly. Is it (8!/6!)*2! or 8!/(!6*!2)?
If it is (8!/6!)*2! Then the solution:
8! = 8x7x6x5x4x3x2x1, 6! = 6x5x4x3x2x1, 2!=2. So, (8!/6!)*2! = 56x2 = 112.

If it is 8!/(6!*2!) The the solution:
8! = 8x7x6x5x4x3x2x1, 6! = 6x5x4x3x2x1, 2!=2. So, 8!/(6!*2!) = 56/2 = 28

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Sreedevi replied | 01/12/2016

The problem is not clear. If we consider the problem as (8!/6!) * 2! we can proceed like this:
8! can be written as 8*7*6!. 6! can be cancelled both in the numerator and denominator. We are left with 8*7*2 ( since 2! = 2*1 =2) = 112.
If we consider the problem as 8!/(6!*2!). we get 28.

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R

Rahul 10/11/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC)), Permutations

A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?

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Y

Yash replied | 22/11/2016

8

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Sarvajeet replied | 06/12/2016

8.

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M

Madhu 10/11/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC)), Permutations

How many 5 - digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?

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Y

Yash replied | 22/11/2016

Rs. 336

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Sarvajeet replied | 06/12/2016

336.

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D

Divya 10/11/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC)), Permutations

How many 3 - digit numbers can be formed from the digits 1, 2, 3, 4, 6 assuming that repetition is not allowed?

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Sarvajeet replied | 02/12/2016

60

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Sanjeev replied | 03/12/2016

5C3*3! = 10*6 = 60.

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A

Ananya 10/11/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC)), Permutations

Define permutation?

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Abhilash replied | 02/12/2016

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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Sarvajeet replied | 06/12/2016

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 permutation.

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P

Parminder 10/11/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC)), Permutations

How many 3- digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?

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Siva replied | 07/12/2016

336.

0 0
K

Kavya replied | 08/12/2016

504.

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K

Komal 10/11/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC)), Permutations

How many words can be formed without repeating the words from the word MONDAY?

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Tapas replied | 04/12/2016

The number of letters in MONDAY is 6 and all are different. So, the number of words that can be formed is !6 = 720.

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Sarvajeet replied | 06/12/2016

6! = 720.

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I

Ishfaque 10/11/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC)), Permutations

Is 3! + 4! = 7!?

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Tapas replied | 28/11/2016

No.
3! + 4! = (1x2x3) + (1x2x3x4) = 6 + 24 = 30
But, 7! = 1x2x3x4x5x6x7 = 5040.
So they are different.

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Sarvajeet replied | 03/12/2016

No.

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R

Ritu 10/11/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC)), Permutations

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?

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Abhilash replied | 30/11/2016

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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Sarvajeet replied | 06/12/2016

20.

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D

Dileep 10/11/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC)), Permutations

Find n, if (n - 1) P (3) : (n) P (4) = 1 : 9?

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Eti replied | 21/11/2016

n = 9

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Sarvajeet replied | 03/12/2016

9

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V

Vaibhav 10/11/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC)), Permutations

Find r, if (5) P ® = 2 * (6) P (r - 1)?

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Sarvajeet replied | 02/12/2016

3

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T

Thamaraikannan 10/11/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC)), Permutations

How many 5 - digit numbers can be constructed using the digits 0 to 9 if each number starts with 76 and no digit appears more than once?

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Sarvajeet replied | 26/11/2016

336

0 0

Gururaj replied | 28/11/2016

8 x 7 x 6
i.e., 336

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N

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Kanwar Kanishk Thakur replied | 11/11/2016

The word FRIDAY has six letters, therefore total words that can be formed = 6! = 720

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Sarvajeet replied | 06/12/2016

6! = 720.

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C

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Johnson replied | 24/10/2016

5!/2!= 60

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Sarvajeet replied | 06/12/2016

5!/2!= 60.

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H

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Prateek replied | 29/10/2016

120

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Sarvajeet replied | 06/12/2016

120.

0 0

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A

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Sarvajeet replied | 06/12/2016

5040.

0 0

Ravi Ranjan replied | 07 Jan

7!

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A

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Santosh replied | 14/10/2016

The sum of all the digits = 0+1+2+3+4+5+6 = 21
Hance any six digit number created with these digit is divisible by 3.
At the leftmost position '0' can not be there.
So total six digits numbers without repeatation of digits :
6 × 6 × 5 × 4 × 3 × 2 = 4320
If repeatation is allowed than this question become more complicated and complexity will be more then the time :-)

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A

Arun Sharma replied | 15/10/2016

Please ignore my previous answer .

Answer is 1920

The number of digits possible with 1,2,3,4,5 and 6 = 6X5X4X3X2X1 = 720

The number of digits possible with 0,1,2,3,4 and 5 = 5X5X4X3X2X1 = 600

The number of digits possible with 0,1,2,4,5 and 6 = 5X5X4X3X2X1 = 600

Total numbers = 720 +600+600 = 1920

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F

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Hanuman replied | 21/10/2016

11*10*11=1210

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Debojyoti Roy replied | 23/10/2016

Captain can be selected in 11 C 1 ways. after selecting Captain ,Vicecaptain can be selested in 10 C 1 ways and afterth is Wicket Keeper can be selected in 9 C 1 ways.
Therefore required no. Of ways= 11C1 * 10C1 * 9C1
=11 * 10 * 9

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V

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Durgesh replied | 12/10/2016

7C3*5C5
=( 7!/3!*(7-3)! ) / 1
= 7*6*5*4! /3!*4!
= 7*5 =35

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Anish replied | 20/10/2016

7C3*5C5=35

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V

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Debojyoti Roy replied | 23/10/2016

Out of 12 consonants 6 consonants can be chosen in 12C6 ways. Out of 4 vowels 3 can be chosen in 4C3 ways. And all these 9 letters can arrange among themselves in 9! Ways.
So required no. Of ways= 12C6 * 4C3 * 9!

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R

Ritu replied | 24/10/2016

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...  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! «less

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R

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Prateek replied | 29/10/2016

35

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S

Sukhada replied | 14/11/2016

35

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S

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R

Ritu replied | 24/10/2016

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

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Prateek replied | 29/10/2016

322560

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