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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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D

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M

Milan Singh replied | 22/10/2016

If X is chosen 9C4, If X is not Chosen 11C5=> 9C4+11C5 = 588

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Prateek replied | 29/10/2016

588

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P

Padma 27/09/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC))

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K

Kiranbabu replied | 01/11/2016

yes

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Meenakshi replied | 02/11/2016

yes

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H

Harsh 09/06/2016 in  Mathematics(Class XI-XII Tuition (PUC))

What is minor in business maths

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K

Kishore 04/06/2016 in  Chemistry(Class XI-XII Tuition (PUC)), Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC))

Need to score more marks in supplementary.. need to write exam in the month July

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Kamal replied | 02/08/2016

contact me

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D

Divya 28/05/2016 in  Mathematics(Class XI-XII Tuition (PUC))

can i get mr.kannana as my professor?

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Praveen Singh replied | 27/09/2016

contect me

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S

Sagar 26/05/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC))

can someone help me to solve this

1 0
A

Abhishek C replied | 15/06/2016

M is 5 and N is -4

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S

Shalini 26/05/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC))

I received 5 recommendations for my search. How can i get more options, suggestions or choices?

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U.K. 01/05/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC))

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C

Chandan replied | 12/06/2016

5

0 0

Chakraborty T replied | 19/06/2016

Answer is 4

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Rajesh replied | 28/05/2016

520

0 0

Bimal replied | 01/06/2016

a+6d= 40 ....(1)
S13 = 13/2 [2a+(13-1)d]
= 13/2*2 [ a+6d] ........(2)
= 13 *40. From eq (1)
' = 520

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A

Ashutosh replied | 12/05/2016

0.20788
exp(i*x) = cos(x) + i*sin(x)
exp(i*Pi/2) = cos(Pi/2) + i*sin(Pi/2) = i
square both the sides now.
i^i = exp(i*i*Pi/2) = exp(-Pi/2) = 0.20788
Interesting fact is that i raised to the i-th power is actually a real number!

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Md Aslam Ali replied | 24/05/2016

we know that e^(ix)= cos(x)+isin(x)
where i=sqrt(-1)
so e^i=cos(1)+isin(1)
=0.5403+0.8415i
so, i*e^i=i*(0.5403+0.8415i)
=0.5403i+0.8415*i^2
=0.5403i-0.8415

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U.K. 01/05/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC))

find the directrix of the parabola x^2 - 4x - 8y + 12 = 0

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Eti replied | 06/05/2016

y=0

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Kamal replied | 03/06/2016

y=-1

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U.K. 01/05/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC))

find the number of words that can be formed out of the letters of the word "ARTICLE" so that the vowels occupy even places ?

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K

Kalidasan replied | 06/05/2016

among 7 places, number of even places = 3, number of odd places = 4
given number of vowels =3 (a, e i) . number of consonants = 4 ( r,t, c l)
number of ways filling 3 even places with 3 vowels = 3! = 6 and number of ways filling 4 odd places with 4 consonants = 4! = 24 therefore total number of ways =( 6).( 24 ) = 144.

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Sahil replied | 07/05/2016

4!*3!=144

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Lakshya 12/04/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC))

I am in 11th and please tell should i take coching classes or tution classes . please reply to as fast as you can.

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N

Neel replied | 15/04/2016

Take one to one tuition class because in coaching class you'll not get enough attention as there will be multiple students....& also most coaching classes ash hefty amount without any quality education. I take both one to one and coaching classes in academy and so I am well aware of both...

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Kamal replied | 03/06/2016

try on your convenience

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Yeswanth 08/04/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC))

How many triangles can be constructed so that lengths of the sides are three consecutive odd integers and the perimeter is less than 1000?

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N

Neel replied | 15/04/2016

Let the sides be x,x+2,x+4 ,x being an odd integer then , x+(x+2)+(x+4)<1000 , i.e, 3*x<994 ,x being an odd integer. Max value of x can 331
Now x can be any odd no from 1 to 329...like 1,3,5,7,..............,331 and for every such x, you will get one triangle..
From arithmetic progression......331=1+(n-1)*2, i.e.
(n-1)*2 = 330
(n-1) = 165
n=166
which...  more»
Let the sides be x,x+2,x+4 ,x being an odd integer then , x+(x+2)+(x+4)<1000 , i.e, 3*x<994 ,x being an odd integer. Max value of x can 331
Now x can be any odd no from 1 to 329...like 1,3,5,7,..............,331 and for every such x, you will get one triangle..
From arithmetic progression......331=1+(n-1)*2, i.e.
(n-1)*2 = 330
(n-1) = 165
n=166
which gives n=166
So 166 possible triangles «less

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Kamal replied | 03/06/2016

166

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