1in that case it will be 9*9*9
some of my friends are confused with percom so i want all here help them out with theory and problems alike......some are starting now...can nishant bhaiya help???
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49well first of all let the 5 iitans sit on x1, x2, x3, x4, x5 respectively and let them sit in this order only...later we will multiply 5! to the result to get the correct answer
so we are writing inequalities about seating the 5 geniuses:P
x1≥1
x2≥x1+1
x3≥x2+1
x4≥x3+1
x5≥x4+1
20≥x5
now as a class 11 student i do not know how to find all possible solutions to such an inequality directly....so i change them to equalities....
x1=1+a
x2=x1+1+b
x3=x2+1+c
x4=x3+1+d
x5=x4+1+e
20=x5+f (a,b,c,d,e,f≥0)
adding the above five equations we get,
20=5+a+b+c+d+e+f
or, a+b+c+d+e+f=15
so the problem is now reduced to finding all possible values of a,b,c,d,e,f
this can be done by puttng 5 pluses between 15 ones
i.e. by 20C5 ways
now we can permute x1, x2, x3,x4,x5 among themselves in 5!ways
so answer is 20C5X5! and thats the answer..........yeeeeeeeeeeeee[1][1][1]
1i can be wrong only if u state that repeatation is allowed but u havent specified it
49nw it is correct....bt it was never said that repitition nt allowed!!!!!!!!:P
3well subhomoy,i guess i was the first person to hear the two rules from nishant sir.
basically the exact rules were rule 1 and rule 2 u have changed it to rule a and rule b.
1# 28 The block letters contain the real question......the rest of the story is the "FLOPPEST EVER" rubbish cooked up by Subhomoy.... he is good at it !!!
49sorry everyone and u dark nightAIR 410 JEE 2011:P answer is 20C5X5!
19P3.sorry it was a careless mistake and thanx for pointing it out
49nishant bhaiya can you please add one of your genius sums???
49a5b6c4=(ab)x(bc)y(ca)z
so x+y=6
y+z=4
z+x=5
but here we re getting fractional valuse of x y and z.....thats odd
1Another Classic :
there are n nos. 1,2,3,4,5,.............,n where n is odd. Find no. of ways of selecting three of these so that they are in AP.
BUT SUBHO...how did u come to know of it (410) ?
3Really... the problem is classic....30 min passed already and I am nowhere....
1One small hint to get you ONE STEP CLOSER :
if a,b,c are in AP,
then a+c = 2b
So actually you are to find the no. of ways of selecting these nos. so that their sum is an even no.
Now Go for the kill !!!!!!!
21this one has been discussed ----check this -http://targetiit.com/iit-jee-forum/posts/ap-12056.html
49now some similar questions:
no of ways to go from A to C and then come back to B when:
a)don't take same route again
b)takes exactly same route again
c)dont take any part of the route he has taken earlier...
49@aveek:not even u r allowed
now i will quote nishant bhaiya:
Permutation and combination has two rules:
a)there is no formula to permutations and combinations
b)donot ever forget the first rule [1][1].......
well permutations and combinations is all about logic and one who gets it gets the right answer.........
there will be always sum who will try to mug up the formula and do the sums but they wont get all correct in a test like iitjee.
now there are some tips for percom......
i am not outright copying nishant bhaiya but he said something like this...
Permutations and combinations are something that can be solved by cooly looking at the sums....4 da first 3 sums posted, they wont require any formula based approach....try to do it logically.....
1There goes subhomoy !!! NISHANT Sir , do you allow your lectures to be outta copyright ??? ;-)
"now i will quote nishant bhaiya:" :-)) ;-)