IPMAT 2020 - Quantitative Aptitude Question 23
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IPMAT 2020 - Quantitative Aptitude
Matrix elements must be distinct
Total Cases = Selection of any 4 elements from 6 and arranging them
= 6C4 × 4!
= (6×5)/2 × 4 × 3 × 2 × 1
= 6 × 5 x× 4 × 3
= 360
Favourable Case = Matrix should be singular i.e. value of determinant is 0
= a × c = b × d = 0 & elements are distinct
So we have 2 cases
Case 1: = 1 × 6 – 2 × 3
= 0
So if we take
a = 1 then c = 6
& b & d take any value 2 & 3
So this give 2 possibilities
a = 6 then c = 1
b & d take values again 2 & 3
So this also gives 2 possibilities
a = 2 then c = 3
b & d take value again 1 & 6
interchanging the values so this gives 2 possibilities
a = 3 then c = 2
b& d take value 1, 6 both & result in 2 possibilities
Total possibilities of distinct matrix case from case 1
= 2 + 2 + 2 + 2
= 8
Case 2: = 2 × 6 – 4 × 3
Similarly as Case 1, Case 2 also result in 8 possibilities
Favourable Cases = 8 + 8
= 16
Probability = (Favourable Cases)/(Total Cases)
= 16/360
= 2/45