Q.

In an examination the maximum marks for each of the three papers are 50 each. Maximum marks for the fourth paper are 100. The number of ways in which the candidate can score 60% marks in aggregate is

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a

110256

b

110556

c

110456

d

None of these

answer is C.

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Detailed Solution

The candidate must score 150 marks.

∴ Required number

= coefficient of x150 in 1+x++x5031+x++x100= coefficient of x150 in 1x511x31x1011x= coefficient of x150 in 1x5131x101(1x)4= coefficient of x150 in 13x51+3x102x1531x101(1x)4

[leaving terms containing powers of x greater than 150]

= coefficient of x150 in (1x)43. coefficient of x99 in (1x)4+3 coefficient of x48 in (1x)4 coefficient of x49 in (1x)4 =153C1503102C99+351C4852C49.=153152151631021011006+351504965251506= 110556

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