Ooooooops hmm how do I explain ?? hehe ... try to take a simple example let us suppose the word in LAMA and we need to make 2 letter words from that. OKAY ??

 

So let us see ... by using the previous approach

 

We will find coefficeint of x² in the expression

    2! (1 + x)²(1 + x + x²/2)

   

    So we find coefficeint of x² in ( 1 + x² + 2.x).(1 + x + x²/2)

 

    The terms corresponding to x² are 2! ( 1 + 2.1 + 1/2)x²

    which is 7.

 

     Now see the normal way the two letter words can be 

AA, AL, AM, LA, LM, MA, ML

 

which are 7 in number so the approach works okay??

 

The expression will change if u make 5 letter words .. in this case u will replace 4! by 5! and find coefficiet of x^5.

 

okay ... cheers

okay let me explain ... first of all do not use that word SIR .. just a kid like u ... may made it earlier to IIT thats it . ok

 

Now about the approach .. see the letter E can appear in a four letter word for one , two or three times.. OKay ?? Now if the letter E appears 3 times then we need to divide by factorial(3) since otherwise we are counting redundant case.

So note that the x^3 has been divided by factorial(3) which is 6.

Similarily for other letters. Try to understand that x^n denotes that the letter appears n times in the word. Now is it clear ???

 

 

Well yes there is one more method where u actually take case and see that whether we have taken all different letters or some letters are alike and so on.

 

However the method described above is simple as u do not need to worry about the cases. The answer will automatically come. DO you understand the approach that has been used above ??

Great work done this a absolutely correct approach !!!!!!!!!!

 

I hope ruhi this helps u !!

hii

First of all thanx a ton for appreciating our effort. Well the pattern if iit keeps changing and it is not advisable to keep that in mind while preparing, what is important is that u should focus on the key concepts and try to understand them, this will help u a lot, believe it did me when i was preparing for JEE.

 

About the books .. well it all depends on how much time u have .. r u a school going student or a dropper ?? Do you go to coaching classes ? Well if u do and do not have much time then going through what is taught in class will be good enough in my view.

 

I hope this helps u

hii

First of all note that the temperature of gas remains constant, so we have

                                P₁V₁ = P₂V₂

 

Now initially the the pressure in atmospheric pressure only, so let it be P

This is also the pressure of the gas.

                    So, P₁V₁ =  PA.(90)   where A is the area of the cylinder

 

When we start pouring the mercury the disc goes down by 32 cm leaving a space of 32 + 10 = 42 cm in the cylinder because the piston was at 90 cm initially.

 

So now the total pressure from outside is P + 42 which is equal to pressure of the gas.   So,  P₂V₂ = ( P + 42 ).A.58

 

                   So using P₁V₁ = P₂V₂

                          we get PA.(90) = ( P + 42 ).A.58

                    and thus solving for P we get P = 76.125 cm of mercury.

Well let the other mass be M where M >> 200.

         Now by balancing the forces at mass of 200kg we get

                        T - 200g = 200a

         Similarily by balancing forces at mass M we get

                        Mg - T  =  Ma

        Now eliminating  acceleration a we get,

                        T = 400Mg / ( 200 + M)

                         Since M >> 200,  200 can be neglected as compared to M

                         So, T = 400Mg/M

                        And thus T = 400g