sjmit4

Prime fators is 3 for first and 5 for the second number so you have to decide which is greater.. hope this helps you to get the answer..

Dude our blessings are always with you. We'll pray for you and one thing you are forgetting in talking about 700 is that it is for average student If you can have good recos and sop that will be awesome and you can be sure of flying to Canada

I think trial and error is the best way to do this kind of problems Even I'll do it the way IIMPUNE has done it

Hey Your score is still above average which is not all that bad can you tell us what all material you went through. Also what were your scores on tests you had taken. Regards, Satyajeet

Can anyone tell me why value of n has to be an odd value?

x^2 + 7x + 1=0 (x+3)(x+4)=0 roots are x=(-3) and x=(-4) { roots are values of variable at which value equation is zero} now for values less than (-4); both (x+3) and (x+4) are negative hence [equation > 0] for values between (-4) and (-3); (x+3) will be negative but (x+4) are positive hence [equation for values more than (-3); both (x+3) and (x+4) are positive hence [equation > 0] so range is x-3 easy way for such questions (quadratic equation) consider following as number line ...........-4.......... -3 ........... values at which the equation becomes zero is -4 and -3 so write zero below it ...........-4.......... -3 ........... .......... (0).........(0)........... now your number line is divided in 3 regions mark first area as positive second as negative and third as positive again from right side ...........-4.......... -3 ........... .......... (0).........(0)........... (+) ve........ (-) ve ....(+) ve. region with +ve mark shows value of equation greater than 0 region with -ve mark shows value of equation less than 0. you can use this if max power of x is even if max power of x is odd then you need to start marking regions as first negative ,second positive and so on from right side. Hope this helps. In case of any confusion feel free to ask.

It is "81" as 18*4.5= 81..

Hey I am able to extract it .. what problem you are having?

n identical things distributed in r groups [(n+r-1)C(r-1)] so what we can do is as everyone has two coins in three groups we are left with 13-2*3= 7 coins now 7 coins in 3 groups [(7+3-1)C(3-1)]=9C2=36

I think it should be 120 as you can choose first male partner in 5 ways, you can choose first female partner in 4 ways, you can choose second male partner in 3 ways, you can choose second female partner in 2 ways. so total ways = 5*4*3*2=120

Its correct

According to me 52*(51C17*34C17) Let me know if you want explaination for the answer.