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# in an exam there are papers of four different subjects three of them have 50 maximum marks 4th paper has maximum marks 100. find no of ways of scoring exactly 60% marks

9 years ago

Dear Piyush,

Concept : For questions involving distribution of objects (here marks) into groups (here it is 4), we use the binomial co-efficinet method.

You should be aware that co-efficient of x^r in (1-x)^(-n) = (n+r-1)C(r) {that is the approach we would be taking}

Here we let marks in 1st subject be x1, in 2nd be x2, in 3rd be x3 and 4th be x4.

The constraint is clearly, x1, x2, x3 lies in the interval [0,50] and x4 lies in the interval [0,100].

with x1 + x2 + x3 + x4 = 150 (which is 60% of the total 250 marks)

Now we need co-efficent of x^150 in ({ 1 + x + x^2 + x^3 + ....... + x^50 }^3)*( 1 + x + x^2 +...... + x^100 )

ie we need co-eff of x^150 in {(1-x^51)^3}*(1-x^101)*{(1-x)^(-4)}

ie we need co-eff of x^150 in { 1 - 3x^51 + 3x^102 - x^153 }*(1-x^101)*{(1-x)^(-4)}

ie we need co-eff of x^150 in [ 1 - x^101 - 3*x^51 + 3*x^102 ]*{(1-x)^(-4)} ----------- (Note that we have multiplied the terms in the first two brackets and have retained only those terms whose power is less than 150. Other terms with power of x greater than 150, would not contribute to the coeff of x^150)

So the req co-eff is: (Let us say that {(1-x)^(-4)} = Z}

(Co-eff of x^150 in Z) - (Co-eff of x^49 in Z) - 3*(Co-eff of x^99 in Z) + 3*(Co-eff of x^48 in Z) ------ (this is obtained say for example, in the second term co-eff of x^150 in x^101*Z, would be co-eff of x^49 in Z)

This would be evaluated using the co-eff of x^r in (1-x)^(-n)...... as mentioned in the beginning.

So the required Ans would be: [ 153(C)150 - 52(C)49 - 3*102(C)99 + 3*51(C)48 ]......

which you need to evaluate.......

Hope that helps!!!

Regards,