Verified Solution • 2025–26 Session moderate

QJEEAdv19Mafe70: JEE Advanced 2019 Mathematics - Class 11 JEEAdv19Mafe70 Solution: Let

Let |X| denote the number of elements in a set X. Let S = {1, 2, 3, 4, 5, 6} be a sample space, where each element is equally likely to occur. If A and B are independent events associated with S, then the number of ordered pairs (A, B) such that 1 ≤\le |B| < |A|, equals .............

Step-by-Step Verified Solution

Given sample space S = {1, 2, 3, 4, 5, 6} and let there are i elements in set A and j elements in set B.

Now, according to information 1 ≤\le j < i ≤\le 6.

When number of element in set B = 1 then number of elements in set A can be 2 or 3 or 4 or 5 or 6. Number of such pairs of A and B in this case

= 6C1[6C2 + 6C3 + 6C4 + 6C5 + 6C6]

When number of element in set B = 2 then number of elements in set A can be 3 or 4 or 5 or 6. Number of such pairs of A and B in this case

= 6C2[ 6C3 + 6C4 + 6C5 + 6C6]

When number of element in set B = 3 then number of elements in set A can be 4 or 5 or 6. Number of such pairs of A and B in this case

= 6C3[ 6C4 + 6C5 + 6C6]

When number of element in set B = 4 then number of elements in set A can be 5 or 6. Number of such pairs of A and B in this case

= 6C4[ 6C5 + 6C6]

When number of element in set B = 5 then number of elements in set A can be 6. Number of such pairs of A and B in this case

= 6C5[ 6C6]

So, total number of ways of choosing sets A and B

= 6C1[6C2 + 6C3 + 6C4 + 6C5 + 6C6]

+ 6C2[ 6C3 + 6C4 + 6C5 + 6C6]

+ 6C3[ 6C4 + 6C5 + 6C6]

+ 6C4[ 6C5 + 6C6]

+ 6C5[ 6C6]

= Sum of all possible products of two terms from

6C1, 6C2, 6C3, ......., 6C6 = p(Assume)

Now we know,

(6C1 + 6C2 + .....+ 6C6)2

= [ (6C1)2 + (6C2)2+ ....+ (6C6)2] + 2[6C16C2 + 6C16C3 + ... +6C16C6

+ 6C26C3 + 6C26C4 + ... +6C26C6+ ....+ 6C56C6]

⇒\Rightarrow (6C1 + 6C2 + .....+ 6C6)2

= [ (6C1)2 + (6C2)2+ ....+ (6C6)2] + 2(p)

⇒\Rightarrow (26 - 6C0)2 = [ 12C6 - (6C0)2 ] + 2p

⇒\Rightarrow p =(26−1)2−(12C6−1)2= {{{{({2^6} - 1)}^2} - ({}^{12}{C_6} - 1)} \over 2}

=(63)2−12!6!6!+12= {{{{(63)}^2} - {{12!} \over {6!6!}} + 1} \over 2}

=3969−924+12=30462=1523= {{3969 - 924 + 1} \over 2} = {{3046} \over 2} = 1523

Note :

(nC0 + nC1 + .....+ nCn)2

= (nC0)2 + (nC1)2+ ....+ (nCn)2 + 2[ nC0nC1 + nC0nC2 + ... +nC0nCn

+ nC1nC2 + nC1nC3 + ... + nC1nCn

+ nC2nC3 + nC2nC4 + ... +nC2nCn+ ....+ nCn - 1nCn]

⇒\Rightarrow (2n)2 = 2nCn + 2(Sum of all possible products of two terms from

nC1, nC2, nC3, ......., nCn)

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