Set Solving - 1

Algebra Level pending

Let there be two subsets $A$ and $B$ of the set $X$ containing $n$ elements.

What is the probability that $A$ and $B$ have the same number of elements? (For $n=2018$ )

If the answer is in the form of $\frac {a!}{(b!)^c2^d}$ , where a,b,c,d are all integers, find $a+b+c+d$ .

The answer is 10092.

1 solution

Parth Sankhe
Nov 17, 2018

The answer is $\frac {^{2n} C_n}{2^{2n}}$ .

Putting n=2018 and expanding $^{2n} C_n$ will give you $a=4036, b=2018, c=2, d=4036$

No. of ways the 2 subsets could have equal number of elements = $ⁿC_0\cdot ⁿC_0 + ⁿC_1\cdot ⁿC_1 + ⁿC_2\cdot ⁿC_2 + .... + ⁿC_n\cdot ⁿC_n = \sum (ⁿC_r)^2=²ⁿC_n$

Total number of subsets possible = $(ⁿC_0+ⁿC_1+ⁿC_2+...+ⁿC_n)(ⁿC_0+ⁿC_1+ⁿC_2+...+ⁿC_n)=(2^n)^2$

Set Solving - 1

The answer is 10092.

1 solution

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