(*args)
return rv
Return product of Permutations [a, b, c, ...] as the Permutation whose
        ith value is a(b(c(i))).

        a, b, c, ... can be Permutation objects or tuples.

        Examples
        ========

        >>> from sympy.combinatorics.permutations import _af_rmul, Permutation
        >>> Permutation.print_cyclic = False

        >>> a, b = [1, 0, 2], [0, 2, 1]
        >>> a = Permutation(a); b = Permutation(b)
        >>> list(Permutation.rmul(a, b))
        [1, 2, 0]
        >>> [a(b(i)) for i in range(3)]
        [1, 2, 0]

        This handles the operands in reverse order compared to the ``*`` operator:

        >>> a = Permutation(a); b = Permutation(b)
        >>> list(a*b)
        [2, 0, 1]
        >>> [b(a(i)) for i in range(3)]
        [2, 0, 1]

        Notes
        =====

        All items in the sequence will be parsed by Permutation as
        necessary as long as the first item is a Permutation:

        >>> Permutation.rmul(a, [0, 2, 1]) == Permutation.rmul(a, b)
        True

        The reverse order of arguments will raise a TypeError.
Return product of Permutations [a, b, c, ...] as the Permutation whose
        ith value is a(b(c(i))).
Return product of Permutations [ a b c ... ] as the Permutation whose ith value is a ( b ( c ( i ))) .
def rmul(*args):
        """
        Return product of Permutations [a, b, c, ...] as the Permutation whose
        ith value is a(b(c(i))).

        a, b, c, ... can be Permutation objects or tuples.

        Examples
        ========

        >>> from sympy.combinatorics.permutations import _af_rmul, Permutation
        >>> Permutation.print_cyclic = False

        >>> a, b = [1, 0, 2], [0, 2, 1]
        >>> a = Permutation(a); b = Permutation(b)
        >>> list(Permutation.rmul(a, b))
        [1, 2, 0]
        >>> [a(b(i)) for i in range(3)]
        [1, 2, 0]

        This handles the operands in reverse order compared to the ``*`` operator:

        >>> a = Permutation(a); b = Permutation(b)
        >>> list(a*b)
        [2, 0, 1]
        >>> [b(a(i)) for i in range(3)]
        [2, 0, 1]

        Notes
        =====

        All items in the sequence will be parsed by Permutation as
        necessary as long as the first item is a Permutation:

        >>> Permutation.rmul(a, [0, 2, 1]) == Permutation.rmul(a, b)
        True

        The reverse order of arguments will raise a TypeError.

        """
        rv = args[0]
        for i in range(1, len(args)):
            rv = args[i]*rv
        return rv
def rmul ( * args ) : rv = args [ 0 ] for i in range ( 1 , len ( args ) ) : rv = args [ i ] * rv return rv