IIT JEE , AIEEE Forum - Mathematics , Algebra

caprio.nups

A person goes in for an examination in which there are four papers with a maximum of 'm' marks from each paper. The number of ways in which one can get '2m' marks is

a. 2m+3 C 3 b. 1/3(m+1)(2m^2 +4m+1)

c. 1/3(m+1)(2m^2 +4m+3) d. none of these

abhijeet_0201

i think the ans is (a)
a+b+c+d=2m
no of sol of this eq is no of ways=(2m+4-1)C(4-1) (this includes the case when he gets 0)
plz plz rate me if my ans is correct. if not plz tell me the mistake

caprio.nups

nopes!!!!
tha ans is C

konichiwa2x

Let

represent the marks scored in each of the exams.

where

Thus, the number of ways the person can obtain '2m' marks is equal to the coefficient of

in the expansion of

i.e, coefficient of

in

= coefficient of

in

= coefficient of

in

(In the expansion of the first term, the higher powers are greater than 2m and hence, you dont need to consider them)

=

Simply and you get,

=

Do ask if any step above is not very clear.

abhijeet_0201

can u tell me the coefficient method.i don't know it.plz tell me

caprio.nups

thanx..

konichiwa2x

well this is the coefficient method if I got u right..

C(n+r-1,r) gives the number of solutions only when

AND

...

Thus the number of solutions is equal to the coefficient of

in the expansion of

= C(4+2m-1,2m)

got it? the maximum value of each of the terms must also equal the value of 'n' to use that formula.

(by the way my name is abhijith too!)

abhijeet_0201

whats n here????

konichiwa2x

sorry for late reply.

n is 4..

hint : C(n+r-1,r) is the number combinations of 'n' objects taken 'r' at a time when each may occur once, twice, or thrice.. or 'r' times in any combination.

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