(3!)!/(3!*3!) = 20, where n! = 1*2*...*n is n-factorial
Proof :
3! =1*2*3=6, (3!)!=6!=1*2*3*4*5*6=720, therefore : (3!)!/(3!*3!) = 720/6*6 =720/36 = 20
20
The first three is 3000 and the second three is worth 300.
Let F = father's age Let S = son's age F = 3S F - 5 = 4(S-5) F - 5 = 4S -20 F = 4S - 15 F = 3S 3S = 4S-15 S = 15, age of son
3 * 3 / ( 3 * 3 ) = 1 but that uses only four 3s, so 33 / ( 3 * 3 * 3 ) = 1 uses five 3s
331/3 = 3,333.333... percent (the 3s can never end)
(3 + 3) / .3
(3+3) / .3
33/3 = 11
20
20 divided by 3 = 6 and 2/3
4=4(3s) 4=12s s= 1/3
The first three is 3000 and the second three is worth 300.
2,560
Let F = father's age Let S = son's age F = 3S F - 5 = 4(S-5) F - 5 = 4S -20 F = 4S - 15 F = 3S 3S = 4S-15 S = 15, age of son
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3 * 3 / ( 3 * 3 ) = 1 but that uses only four 3s, so 33 / ( 3 * 3 * 3 ) = 1 uses five 3s
There are three thirds (1/3s) in every whole, so there are six thirds in two wholes.