[Legal counting forum game] Counting with four fours

Is that calculator telling you that the square root of 4 is 1.41...?
Also, 49 = 4!+4!+(4/4)

I've seen that before, but I forgot what causes it.

And now I'm finally going to play too!
4P4 + 4! + √4 (= 50)

Isn't (4P4) just (4!)?

Anyway,
((4!-sqrt(4))/.4)-4 = 51

52 = 4!/0.4 - 4 - 4

Yeah. xPx = x!. But 4P4 uses an additional 4.

Edit: I'll try to do 53.
Another edit: Didn't succeed.

I beleve he just did sqrt(sqrt(4)) but only the down one is shown.

4P4 = 4! yes, it just is a way to use an extra four if you have used only three. but the whole function equuals

nPk = n! / ((n-k)!) it is used when you choose k things out of n things and the order you do that in matters.

This is an a little bit cheety one, there should be better, but for now, I guess it is good, as square root also is allowed...
= 53

You could also use nPr differently than just 4P4, eg. if I would do ((4+4)p(4)) it would give 1680, since there are 1680 ways you could arrange 4 random objects out of a 8 object array.

54: ((4 + sqrt(4))! / 4!) + 4!

^ 0.5 and ^ 2 aren't that equal, though. But fine, your rules...

but when you would write 0.5 as 1/2 then it looks alot more alike already

However I do agree that sqrt seems a bit more 'allowed' than ^2, since you can write a sqrt without using a number (in this case 2), because I dont think a 3rd power root is allowed either (since you would have to write a 3 then).

Well... if you look at it from this way, yes, it seems like that, but this was because the cutoff from the rest of the number was just right. But what it actually said was




EDIT: grrr I got 54 and 56...... but no 55! Let's try some more

Cool, but that's √4 x 4 4s, not 4.

EDIT: this is 55 with 5 4s, still 1 too many...

I said it was a little cheety, but I hhad some serious troubles withh finding it, and you didn't even find it, so I thought it was fine in thaht sense...
also: I found a way to do 55 withhout any cheety buisness, but I cannot post again, if no one can come up a way, I'll post it. (I also would have to post a picture of my calculator, I can't find intergrations in the word thing...
EDIT: It usues five fours aswell, stupid idiot I am, never mind me, keep on searching

I know this probably isn't *entirely* legal, but we're stuck.
55 = 4!/0.4 - 4 - √√√√√√√√√√√√4

since y'all've been stuck on these:
((4!-sqrt(4)/.4)-sqrt(4) = 53
4!/.4-4-sqrt(4) = 54
(4P4-sqrt(4))/.4 = 55 Hey I used 4P4 too

Now
4!+4!+4+4 = 56

57 = (4!-sqrt(4))/.4 + sqrt(4)

(4P4)/.4-sqrt(4) = 58

59 = 4!/0.4 - (4/4)

4!/.4+4-4

60 = 4!/.4 +4 -4

Ninja's but:

Bothh technically illegal posts due to two posts being in between, but due to the way this thread has been working so far, I'm going to change that rule up to be one in between, as that would make it usefull again

61 = 4!/0.4 + (4/4)