MadeEasy Test Series: Numerical Ability - Permutation And Combinations

Question

A permutation group 'P' is defined as a one-one mapping of non-empty finite set X, onto itself i.e.,

P: X-> X.

where X={1,2,3,...,n}

S is the set of all permutation 'P' on X.

$P1=\begin{pmatrix} 1 &2 &3 &... &n \\ P1(1)&P1(2) &P1(3) &... &P1(n) \end{pmatrix}$,$P2=\begin{pmatrix} 1 &2 &3 &... &n \\ P2(1)&P2(2) &P2(3) &... &P2(n) \end{pmatrix}$ and so on.

what is the cardinality of power set of S _______? Given n=3.

Comments

64 ??

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Yes it's 64 but how?

Answer 1

n=3.

One-one mapping from the set to itself. Hence, number of one-one functions = n! =6. P has 6 mappings.

Now, S is the power set of P, i.e. 2^6=64. So S has 64 elements.

Comments

what does this line mean?