106suppose x+y+z = n (n is non-negative)
x can assume any value from 0 to n
similarly y and z
If we represent the values x can take as the coeff of x as x0,x1,x2 ...
Then try to find the coeff of xn in
(x0+x1+x2+..+xn)(x0+x1+x2+..+xn)(x0+x1+x2+..+xn)
If you pick 1 from the first bracket and 2 from the second then u have to select n-3 from the third
Since the terms are multiplied, the coefficients get added to get n
So finding the coeff of xn is essentially finding the various no. of ways that can be there such that x+y+z=n
Now coeff of xn in (x0+x1+x2+..+xn)3 is the same as coeff of xn in (x0+x1+x2+..)3 upto infinite because after the coeff becomes greater than n there wont be any values of x^n
So it is equivalent to coeff of xn in (1-x)-3
1number of positive solns (non 0) of a1+a2+a3....an=k
is same as the number of ways in which i can place k similar balls in n different boxes, such that no box is empty..
for simplicity case take 3 variables, x, z, and y.
so x+y+z=9 (say),
then one of the cases could be 1,1,7
the other 2,1,6 and say 7,1,1 so these make up 3 cases so we need the total count of the number of such possible cases,
which is the coefficient of x^9 in (x+x2+x3+...x^9)(x+x2+x3+...x^9)(x+x2+x3+...x^9)
here we see that the minimum power of x u can pick up from each bracket is 1, and the max u can pick up is 9 (which will not be possible)
so that way it wud be the coeff. of x^9 in (x+x2...x9)^3
similarly, think of a total of k balls, and n boxes,
u wud get coeff of x^k in (x+x2...xk)^n
if u think along similar lines, u wud also be able to derive the permutations case..
cheers!!
1
gordo
·2009-12-14 23:14:34
@ppl 150 is the right answer
@xyz,
when u have n different objects and u have to put them in m different boxes such that no box goes empty, remember this simple derivation,
the total number of ways by which u can do this shud be
the coefficient of x^n, in
n!(x+x2/2!+x3/3!...+x^n/n!)(x+x2/2!+x3/3!...+x^n/n!).....m times
reason being that to get the answer we need to find summation
n!/x!y!z! where x+y+z=n
which is the fall out of the above bracket expansion for the coeff. of x^n
now it wouldnt hurt to go any further than x^n in (x+x2/2!+x3/3!...+x^n/n!) as such cases would by themselves be eradicated in the final answer,
so y not sum take it to an infinite series, hence
coeff. of x^n in n!(x+x2/2!+x3/3!...+x^n/n!+....)*(x+x2/2!+x3/3!...+x^n/n!+...)mtimes
=coeff of x^n in
n!*(x+x2/2!+x3/3!.....∞)^n
=coeff of x^n in n!(e^x-1)^m
but the general observation has been that they never ask the direct application of these formulae in jee, i mean u can use them, but it would be simpler and less time consuming if u used the direct approach
cheers!!
1sry .....printing mistake in the second part!It will be.....5C2 x 3C2 instead of .5C3 x 3C2
1
Nikhil Kaushik
·2009-12-14 22:11:35
this is in accordance with 5 = (3 + 1 + 1) = (2 + 2 + 1)
1
Nikhil Kaushik
·2009-12-14 22:10:12
the exact solution is 5C3 x 3! + 5C3 x 3C2 x 3!/2 = 6- + 90 = 150. :)
21@aveek its jus total ways of arrangment of this (2,2,1)---3!/2!
21
eragon24 _Retired
·2009-12-14 08:21:36
its simple
since each box can hold five balls
no of ways in which balls can be distributed So tat none is empty r (2,2,1) (3,1,1)
(5C23C21C1+5C32C11C1)
=50
so total ways 20+30=50
however der r exactly 3 such arrangments possible so total ways r 50x3=150
1yeh IIT-1981 ka question hai Ans : 150
11I think it's 150....ye fiitjee qsn hai.
21
eragon24 _Retired
·2009-12-14 09:17:15
check once more ........it shud be 300 only
11
Devil
·2009-12-14 09:16:56
Keep this in mind.....
The number of ways in which n different things can be distributed in r different groups is coefficient of xn is n!(ex-1)r....however null groups are disallowed.
1i think it will be lyk this
5C2*5C4 + 5C1*5C3*2C1
1
Bicchuram Aveek
·2009-12-14 08:40:37
@Arshad : BOXES ARE DIFFERENT : IT"S MENTIONED
1
Bicchuram Aveek
·2009-12-14 08:28:16
Answer's 150.
the entire lot divided by 2! ..... but why ?
