We are going to be having our son in a few days and we can't decide on James Liam Fredrick. or James William Fredrick.
Which sounds better, Liam or William?
My ex's sister thinks that I am using William to name the baby (not my ex's) after his late father. That's why I'm trying to change it to Liam.
2007-12-22 04:05:20 · 26 answers · asked by Anonymous in Pregnancy & Parenting ➔ Baby Names
I wouldn't let your ex's sister change your mind on William. Anyway, I think James William Fredrick goes very well together. Congratulations on not only the new arrival, but good choice in name. Happy holidays.
2007-12-22 04:19:12 · answer #1 · answered by LMo 2 · 0⤊ 1⤋
James Liam Fredrick
2007-12-22 14:12:12 · answer #2 · answered by Keligh P 6 · 0⤊ 0⤋
Personally I prefer James William Fredrick, but James Liam is good also. If you're having doubts about James William then maybe your mind is made up already.
However, if you decide that you like James William more, it really shouldn't bother you too much what your ex's sister thinks.
Try them both out with your surname and see which one sounds best.
A Christmas baby!! Congrats! :-)
2007-12-22 12:21:57 · answer #3 · answered by So_Long_Marianne 6 · 0⤊ 0⤋
I like James Liam Fredrick
2007-12-22 12:20:02 · answer #4 · answered by Anonymous · 1⤊ 0⤋
James Liam Fredrick sounds nice.
2007-12-22 12:16:35 · answer #5 · answered by healthnut 5 · 1⤊ 0⤋
I like James William better than James Liam.
2007-12-22 12:13:33 · answer #6 · answered by goodbye 4 · 0⤊ 1⤋
I think William sounds better in this case but I love the name Liam. What about Liam James Fredrick?
2007-12-22 12:09:30 · answer #7 · answered by (S.)Born in the spring!☮ 5 · 1⤊ 2⤋
William sounds better. But if there's a controversy, change it.What about Liam James? Then there will be no questions.
2007-12-23 19:18:12 · answer #8 · answered by paula r 7 · 0⤊ 0⤋
I think James William Fredrick sounds alot better together,
2007-12-22 12:15:14 · answer #9 · answered by ♥AsH♥ 5 · 0⤊ 1⤋
Liam
2007-12-22 13:21:38 · answer #10 · answered by HelpMe 2 · 0⤊ 0⤋