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Greatdreams

In an entrance examination there are 2 papers each of 50 marks and 1 paper of 100 marks.A candidate will qualify to take admission if he secures at least 150 marks in aggregate,securing at least 50 marks in the last paper.In how many different ways can he secure marks in 3 papers to qualify?

Answer provided is 655189.

Someone solve this problem for me.

Source : A.Dasgupta , Permutation and Combination, Pg A-173,174 of my book.,Subjective Exercises ~ Problem no.114.

LAMPARD

I am getting a different answer...but i am not sure whether its correct..

We will have to take cases.Let us assume he gets exactly 150 in the test.

And he also gets 50 in the 3rd paper.So,in the remaining 2 papers,he will have to get 50 each(for total to be 150).So,this can be obtained in 1 way,i.e,50+50+50.

Now,suppose he gets 51 in the 3rd paper.So,he will have to get a total of 99 in the first 2 papers.(for total of 150).So,this can be obtained in 2 ways,i.e,49+50+51 and 50+49+51.

Now,suppose he gets 52 in the 3rd paper.So,he will have to get a total of 98 in the first 2 papers.This is possible in 3 ways,i.e,49+49+52,48+50+52 and 50+48+52.

Thus,we can see that as the marks of the 3rd paper are increased by 1,the number of cases also increase by 1.Thus,for him to obtain 150 marks in the test,the number of ways are 1+2+3+..+50.

But the sum isnt over yet.He can also score more than 151.Suppose he scores 151,then we will again take cases.
Suppose he scores 50 in the 3rd paper.Then the maximum total he can get is 150 as the max. marks are 50 in the previous 2 papers.So,this case is eliminated.

Suppose he scores 51in 3rd paper,then he will have to get a total of 100 in the first 2 papers which is possible in only 1 way.

Similarly,taking cases,you will see that the number of cases goes on increasing by 1 with 1 increasing mark in the 3rd paper.

So,total number of ways of obtaining a total of 151 is 1+2+3+..+49.(Note:Here,50 wont be there as 1st possibility is neglected.)

Thus,you can see that with a 1 mark increase in the aggregate,there is a decrease in the number of cases in the last term.Number of ways of getting 152 will be 1+2+3+...+48.

Thus,the total number of ways of getting any mark between 150 and 200 is

1+2+3+...+50 +1+2+3+...+49 +1+2+3+...+48 +1........+1.
The last +1 is the number of ways of getting 200,i.e,50+50+100.
Thus,the total number ways(say,S) can be written as

S=50(1)+49(2)+48(3)+47(4)....+1(50).
Thus,S=_[r=1]

^[r=50] r(51-r)=_[r=1]

^[r=50] 51r -_[r=1]

^[r=50] r².
Then using the formula for summation,you can get the answer.But it is coming out to be something else.I think i may be wrong somewhere.

Greatdreams

Thank you for so much of your time.....i also got a different answer....

.Generally their answers are never wrong.

LAMPARD

But is my method correct?Did you also get the same answer?

nadeemoidu

The given answer is obviously wrong. the total no. of ways of getting marks ( without considering any constraints) is 51x51x101.
This is much less than 650000

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