CAN YOU TRY OUT SOLVING THIS!
2006-08-25 21:40:03 · 15 answers · asked by praveena k 1 in Science & Mathematics ➔ Mathematics
1! = 1
2! = 2x1 = 2
3! = 3x2x1 = 6
4! = 4x3x2x1 = 24
5! = 5x4x3x2x1 = 120
6! = 6x5x4x3x2x1 = 720
from the list you can see that 6x120 = 720 therefore 3!*5! = 6!
but since 3 factorials are needed at the left side of the equation, then answer is 1!*3!*5! = 6!
2006-08-25 22:35:50 · answer #1 · answered by mei mei 4 · 1⤊ 0⤋
1 + 2 + 3 = 6
2006-08-26 06:34:41 · answer #2 · answered by SAMUEL D 7 · 0⤊ 0⤋
a = b = c = d = 1
2006-08-26 04:43:54 · answer #3 · answered by Catsmoking 1 · 0⤊ 0⤋
SOLUTION
a = 1
b = 1
c = any natural number
d = c
ANOTHER SOLUTION
a = 1
b = any natural number
c = b! - 1
d = b!
from the second solution I can give infinitely many solutions a,b,c and d where a, b, c and d are all different natural number.
Here are some:
(a,b,c,d)
(1, 3, 5, 6)
(1, 4, 23, 24)
(1, 5, 119, 120)
(1, 6, 719, 720)
(1, 7, 5039, 5040)
(1, 8, 40319, 40320)
(1, 9, 362879, 362880)
(1, 10, 3628799, 3628800)
...
...
^_^
2006-08-26 07:12:52 · answer #4 · answered by kevin! 5 · 1⤊ 0⤋
I have seen some answers that don't make sense.
1!3!5!=6! does work.
since 3! is 3*2*1=6 and 5! = 5*4*3*2*1
so 5!*6 = 6*5*4*3*2*1 = 6!
2006-08-26 05:26:10 · answer #5 · answered by TRE 3 · 1⤊ 0⤋
1!*2!*3!=4!
2006-08-26 04:56:42 · answer #6 · answered by raj 1 · 0⤊ 0⤋
3!*4!*5!=6!
2006-08-26 04:56:11 · answer #7 · answered by shagun_d_cool 1 · 0⤊ 0⤋
put a=b=c=d=1
2006-08-26 04:48:47 · answer #8 · answered by light feather 4 · 0⤊ 0⤋
a=b=c=d=1
or
a=b=c=d=0 (since 0!=1)
If b=c=0or1, then a!=d!
done!!!
2006-08-26 06:13:11 · answer #9 · answered by Anonymous · 0⤊ 0⤋
if a,b,c,d can all be the same thing, then a=b=c=d=1, or a=b=c=d=0
2006-08-26 05:12:23 · answer #10 · answered by dudemanyeah 2 · 0⤊ 1⤋