Numbers 1,2,3,4 from 100 to 150

Following on from last month's post about achieving the numbers from 1 to 50 (and then from 1 to 100) using just the digits 1,2,3,4 and the maths operators + - / *  and powers and factorials, I've gone a step further (and possibly a step too far) - from 101 to 150.  This is proving far more difficult - as you would expect - and so far I've only managed around 30 of the 50.  If the numbers below are blank, it's because I haven't solved them yet :-)

Major thanks to Denis and to lookagain from the Free Math Help Forum for their many solutions and for helping me fill the gaps that I had.  Denis's solutions are in purple, lookagain's are indicated in blue.  You'll see that in two cases, we've had to use the FLOOR function (i.e. the result of an operation is rounded down to the nearest integer) - we're trying to avoid using this where possible.  If you have any better solutions, please let me know!

101 = (4! / .3) + 21

102 = 3! * (21-4)

103 = 4√.1∗.2 + 3

104 = (4! + 2) * (3+1)

105 = 21 * [3 + SQRT(4)]

106 = (3+2)! - 14

107 = 4!.2−13

108 = 4 * (3 ^ (2 + 1))

109 =  4!−1.2−3!

110 = (31 + 4!) * 2

111= (1+4)! - 3^2

112 = (1+3!) *4^2

113 = (4! / .2) -1 - 3!

114 = (1+4)! – (3 * 2)

115 = (1+4)! – (3 + 2)

116 = (3+2)! – (1 * 4)

117 = (3+2)! – (4 - 1)

118 = 124 - 3!

119 = (4+1)! – (3 - 2)

120 = (4+1)! * (3 - 2)

121 = (4+1)! + (3 - 2)

122 = (3! * 21) – 4

123 = (42 – 1) * 3

124 = (3+2)! + (1 * 4)

125 = (1+4)! + 3 + 2

126 = 42 * 3 * 1

127 = (42 * 3) + 1

128 = 1 *(2 ^ (3+4))

129 = (42 + 1) * 3

130 = (3! * 21) + 4

131 = 1+4!.2+3!

132 = (31 + 2) * 4

133 = 4! / .2  + 13

134 = (3+2)! + 14

135 = (21+4!) * 3

136 = 132 + 4

137 = FLOOR (412/3)

138 = (24 - 1) * 3!

139 = 142 - 3  

140 = (1 + 3 + 4!) / .2

141 = (3! * 4!) – (2 + 1)

142 = (3! * 4!) – (2 * 1)

143 = (3! * 4!)  – (2 - 1)

144 = (3 * 4) ^ (2 * 1) = (3 * 2)! / (4 + 1)

145 = (13^2) - 4!

146 = (3! * 4!) + (2 * 1)

147 = (3! * 4!) + 2 + 1

148 = 142 + 3!

149 = (4! + 3!) /0.2  - 1

150 = 3! * (21 + 4)