IIT JEE , AIEEE Forum - Mathematics , Algebra

raja987654321

The number of distinct quadratic equations of the type Ax²+Bx+C=0 that can be formed when A,B,C are selected from {1,2,3,4,5,6} is:

(i) 216 (ii) 198 (iii) 181 (iv)120

neeraj_agarwal_1990

no. of distinct 3 digits from {1,2,3,4,5,6} = 6C3

so no. of distinct equations =6C3 . 3!=120 (A,B,C are distinct )

and 6 other equations can be formed when A=B=C
but if we have to pick them without replacement...then answer should be 120 according to me....

if its with replacement then answer should be 126.....

richa_dpsvk

number of equations should be 6P3 = 120 according to me

raja987654321

sry answers dont conform ans is 181 shall rate anybody solving it

nadeemoidu

A,B and C can have values 1 to 6 [ repetition is allowed]

so total no. of values possible = 6* 6* 6 = 216

now many cases will give the same equation . e.g. x² + x +1=0 and
2x²+2x + 2 =0 are same.

So let us find the total no. of such repetitions

let the equation be denoted by A-B-C
1-1-1 can be formed in 6 ways.
Here 5 are extra.

1-1-2 , 2-2-4 and 3-3-6 are same.
So there are 2 extra cases and it can be permuted in 3 ways each
total = 2x3 =6

Similarly 1-2-2 has      6 ways

1-1-3 has 3 ways

1-3-3 has 3 ways
1-2-3   . . . 6 ways
2-2-3   . . . 3 ways
2-3-3   . . . 3 ways

total of all extra equations = 5+6+3+6+3+6+3+3 = 35

Reqd. answer = 216 - 35 = 181

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