Let X be a set containing 6 elements and P X be its power set- The sets A and B are picked from P X - If n A - n B and A - B- then total number of ordered pair A - B is -Note - n A represents number of elements in set A -A- 660- 760- 960D- 860

The correct option is D 860Required number of ways =( ^6C_1)^2 ^6C_1+( ^6C_2)^2 ^6C_2+…+( ^6C_5)^2 ^6C_5 =∑_r=1^5( ^6C_r)^2 ∑_r=1^5( ^6C_r) =[12 !(6 !)^2 2] ...

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Standard XII

Mathematics

Combination

Let X be a se...

Question

Let $X$ be a set containing $6$ elements and $P (X)$ be its power set. The sets $A$ and $B$ are picked from $P (X) .$ If $n (A) = n (B)$ and $A \neq B,$ then total number of ordered pair $(A, B)$ is
[Note $: n (A)$ represents number of elements in set $A$ ]

660

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760

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860

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960

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Solution

The correct option is C $860$
Required number of ways
$= (^{6} C_{1})^{2} -^{6} C_{1} + (^{6} C_{2})^{2} -^{6} C_{2} + \dots + (^{6} C_{5})^{2} -^{6} C_{5}$
$= 5 \sum r = 1 (^{6} C_{r})^{2} - 5 \sum r = 1 (^{6} C_{r})$
$= [\frac{12!}{(6!)^{2}} - 2] - [2^{6} - 2]$
$= \frac{12!}{(6!)^{2}} - 2^{6} (∵ n \sum r = 0^{n} C_{r} = 2^{n}, n \sum r = 0 (^{n} C_{r})^{2} = \frac{(2 n)!}{n!})$
$= 924 - 64$
$= 860$

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Let X be a set containing 6 elements and P(X) be its power set. The sets A and B are picked from P(X). If n(A)=n(B) and A≠B, then total number of ordered pair (A,B) is [Note :n(A) represents number of elements in set A]

The correct option is C 860Required number of ways =( 6C1)2− 6C1+( 6C2)2− 6C2+…+( 6C5)2− 6C5 =5∑r=1( 6Cr)2−5∑r=1( 6Cr) =[12 !(6 !)2−2]−[26−2] =12 !(6 !)2−26 (∵n∑r=0 nCr=2n,n∑r=0( nCr)2=(2n)!n !) =924−64 =860

Let $X$ be a set containing $6$ elements and $P (X)$ be its power set. The sets $A$ and $B$ are picked from $P (X) .$ If $n (A) = n (B)$ and $A \neq B,$ then total number of ordered pair $(A, B)$ is
[Note $: n (A)$ represents number of elements in set $A$ ]