1) Different postition of the arms result in different signals, thats understood .So total no. of signals are 4^N, where N is the total no. of arms. But in these selections, there is a combination which has all the arms at rest, thats not a signal. So 4^N -1 = 1023 => N =5

2)Working on this!.

@ little genius: The arrangement can be like this also:

BBGBBGBBGG, in this no boy is between two girls.

3) There are 3 vowels : A,E,A. Not changing the relative ordering means that the first place must be occupied by a vowel, the second and third place by consonants, the fourth place by a vowel, the next two places by consonants and the last place by a vowel.

That gives 4! * (3! /2) , 4 places for consonants and 3 for vowels in which A is repeated once.

Maybe you can give us a hint ? The two terms seem so much unrelated? What should we look for in such problems?

oO I found my mistake. I wasnt counting the no. of handshakes of an American wife and the American husbands except her own, I just thought American wives dont shake hands with the American males. :\

So that adds 4*5 = 20 to the answer and I get 135 handshakes.

computer 001 , you forgot the cases where the Americans and Indians shake hands among themselves. So that adds 4* 5C2 to your answer.

What you actually needed was something else so my answer is right but to know what you needed see this: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=212489

Q) In an exciting finish to an one-day cricket match between India and Pakistan, the Indians require 10 runs in the last 3 balls to win. If any one of the scores 0,1,2,3,4,5,6 can be made from a ball and no wides or no-balls are bowled then in how many different sequences can the batsmen make exactly 10 runs?

My attempt:

I found the coeffecient of x^10 in the expansion of ( 1 + x + x^2 .... x^6)( 1 + x + x^2 .... x^6)(x + x^2 ..... x^6), which comes out to be 33. But the answer is 21.

Q) 5 Indian and 5 American couples meet at a party and shake hands. If no wife shakes hands with her husband and no Indian wife shakes hands with a male, then find the number of handshakes that take place in the party.

I got the answer 115 by simple combinations and inclusion-exclusion but answer is 135.

Both are AD Gupta problems hence even though I am pretty sure of my answers still wanna confirm.

Apply expansion to all the terms :)

Answer is 135. Still you can tell me your logic.

There are 5 eligible Punjabi grooms of which 3 know Bengali and 5 eligible Bengali grooms of which 2 know Punjabi. There are 5 eligible Punjabi brides and 5 eligible 5 eligible Bengali brides. If an eligible groom is agreeable to marry a girl of his community or knowing her language and brides have no choice, in how many different ways 10 couples can be formed?

Srry about my solution then, I did have a doubt in that. I will get back to you with a solution ASAP. But its strange that the answer is less than 1. :\

x^2 = tan@

2xdx = sec^2@d@

Integrand reduces to (sec^2@ * sec^2@)d@/2tan@(1 +  tan^3@)

 tan@ = t 

tan@ = (1+t)^2 dt / 2t(1+t^3)

 Next partial fractions.

Take the larger of the two no.s in the series outside the root. So equation becomes

1+ 2 * nth root {1 +(1/2)^n} + 3 * nth root {1 + (2/3)^n}...................m * nth root {1 + (m-1/m)^n}

Lim n->inf =  1+2 + 3....+m

= m(m+1)/2

Lim m-> inf gives the answer infinity. Is it right?

OK so everything basically means that we are always below the point where we started. A good question and nicely answered by astatine,computer001and feynmann.

Hmm, since superman is superman so ofc he can do anything but the thing is he is carrying a helicopter(I wonder why) which has mass and this will create any diff if any.

But I do think superman(not you :p) can do what you claim.Ofc I am seeing this question as purely ideal situation i.e no air friction , no energy letdowns for superman n all xD.

Ofc I am wrong, but looking forward to your answer anchit.

Yes I got something like that, the book definitely has given the wrong options and answer. :\

Can you try the other ones , especially the fourth one, for which the solution given in the book seems complete absurd.

BUMP BUMP