IIT JEE , AIEEE Forum - Mathematics , Algebra

neeraj_agarwal_1990

A is a set containing elements.A subset P of A is choosen. The set A is reconstructed by replacing the elements of P. A subset Q of A. is again choosen. The number of ways of choosing P and Q so that P

Q contains exactly two
elements is:

Ans. ^-nC₂.3 ^_n-2

rik_mad

hi neeRaj,I CUDN'T UNDERSTAND UR ANSWER VERYWELL BUT THIS IS WAT I GOT WHICH IS SIMILAR TO UR QUERY

Firstly we know that minimum 2 elements must be there in subset P

SELECT 2 ELEMENTS FROM N ELEMENTS OF SET 'A' FOR A TWO MEMBERED SUBSET P

I.E.

^NC₂

(1) NOW Q MUST CONTAIN THE TWO SELECTED ITEMS

AND IN ADDITION IT CAN CONTAIN EVEN MORE

THIS CAN BE DONE IN ^N-2C₀+ ^N-2C₁+ ^N-2C₂+.........+ ^N-2C_N-2 = 2^N-2

SO NUMBER OF WAYS P

Q = 2 ITEMS FOR P WITH TWO ELEMENTS IS

^NC₂(2^N-2)

NOW FOR THREE ELEMENTS IN P SELECTION CAN BE DONE ^NC₃

OUT OF WHICH ANY TWO ELEMENTS CAN BE SELECTED TO BE COMMON IN ³C₂WAYS

SO FOR N(P) = 3 NUMBER OF FAVOURABLE COMBINATIONS BECOMES

^NC₃(2^N-3)(³C₂)

SO THE SUMMATION BECOMES

_{[K=2 ]}

^{[ N]} ^NC_K(^KC₂)(2^N-K)

_{K=2 ]}^{[ N]} N(N-1)/K(K-1) ^N-2C_K-2(K(K-2)/2) {2^N-K)}

_{K=2 ]}^{[ N]} (N(N-1) ^N-2C_K-2 {2^N-K)})/2

N(N-1)/2 _{K=2 ]}^{[ N] N-2}C_K-2 {2^N-K)}

CLEARLY THE SECOND TERM IS A BINOMIAL EXPANSION OF

(1+ 2)^N-2

SO THE SOLUTION BECOMES

N(N-1)/2 (3)^N-2

^NC₂(3)^N-2.

PHEW !! HOPE IT'S RIGHT

deepak_agarwal

rik mad has produced a wonderful soln...kudos to u maan!!!

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