IIT JEE , AIEEE Forum - Mathematics , Algebra

shailesh_45

1] in an examination the maximum marks for each of 3 papers is n and 2n for the fourth paper .prove that the no ways in which a candidate can get 3n marks is ( n + 1)( 5n^2 + 10n + 6 ) /6

2] there are n persons sitting around a table . prove that the no of different ways in which 3 persons can be selected so that no 2 of them are neighbours

3]there aer 8 line segments in cm 1, 2 , .... 8 how many quadrilaterals can be made with these line segments if circles can be inscribed in these quadrilaterals made ?

feynmann

1> is easy

it is just the coeff of x^3n in the expansion of

( 1+ x + x^2 + ... + x^n ) ^3 ( 1 + x + x^2 +...... + x^2n )

Both are GP series so we can find the coeff very easily

mukundmadhav

2nd is nC3- (n-1)X(n-4) - (n-2)

shailesh_45

madhav could u please elaborate how did u get the ans , please explain
thank u sir

RyuAmakusa

let the no. of persons between any 2persons be a,b,c now

a+b+c=n-3 where 1<a,b,c<n-5 now u can use multinomial theorm

mukundmadhav

nC3 is the total number of ways of selecting three people.. (n-1)(n-4) is the number of ways of selecting two people who're together and one person who is not with them. Draw a rough diagram, for around 6-7 ppl, it'll be clear
n-2 is the number of ways of selecting three people sitting together

mukundmadhav

I can attempt the third, but I'm not sure of this answer. I think a circle can be inscribed only if the quadrilateral is convex, i.e it satisfies the triangle inequality, i.e. a +b+c>d
Number of quads not satisfying this are
1,2,3,6
same with 7 and 8 instead of 6
1,2,4,7
1,2,4,8
1,2,5,8
So 6 unfavourable combinations
So total number of fav combinations are 8C4-6

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