Distinct Values that can be Created An expression is formed using the numbers 7, 16, 25, and 27 according to the following rules.
$\bullet$ Each of the four numbers is used exactly once.
$\bullet$ The four numbers may be used in any order.
$\bullet$ Exactly three operations are used; each one is either $+$ or $\times$.
$\bullet$ An unlimited number of parentheses may be used.
No two distinct expressions have the same simplified value. The two expressions below are not distinct, and therefore must be counted as only one value. What is the greatest number of distinct values, including the one below, that can be obtained when building expressions following these rules?
$(7+16+27) \times 25 = 25 \times (27+7+16)$
With one operation $\cdot$ you can form recursively one pair $(x\cdot x)$, two triples $((x\cdot x)\cdot x)$, $(x\cdot(x\cdot x))$, and five quadruples
$$(((x\cdot x)\cdot x)\cdot x),\quad ((x\cdot(x\cdot x))\cdot x),\quad((x\cdot x)\cdot(x\cdot x)),\quad(x\cdot((x\cdot x)\cdot x)),\quad(x\cdot(x\cdot(x\cdot x)))\ .$$
In each of these quadruples we can replace the single  $\cdot$ by one of the two $*$ or $+$ in eight ways, and the four ixes by a permutation of $7$, $16$, $25$, $27$ in $24$ ways. It follows that there are in all $5\cdot8\cdot 24=960$ ways to do the computation. At the end there result only $52$ different values, namely
$$75, 164, 218, 230, 314, 326, 380, 434, 464, 476, 539, 569, 589, 602, \
607, 646, 698, 787, 839, 866, 880, 944, 1082, 1114, 1196, 1250, 1296, \
1376, 1394, 2827, 2989, 3049, 3199, 3232, 3424, 3475, 3699, 4741, \
4837, 5125, 5157, 5824, 7525, 7749, 10807, 10912, 10975, 10989, \
13600, 13824, 15525, 75600\ .$$
This question may be very interesting. Firstly, we can use a method similar to stars and bars to solve the problem. There are 7 places where 4 numbers and 3 plus or multiply may be used. One case is that all 4 are added. First, when 3 are added and the last is multiplied to all. It contains 4 cases. Then, when two are added and rest are multiplied to it. It contains 6 cases.Then, when 2 are added and multiplied by third and last is added. It contains 6*2=12 cases. Then, 2are added multiply by 2 added. It contains 12 cases. Now all parenthesis are used. Left are without brackets. Then, when two are added and two are multiplied. It contains 6 cases. Then, 2 are multiplied added by 2 multiplied. It contains 6 cases. Then 3 are multiplied and left is added. This contains 4 cases. Last case is when all are multiplied.
So, total number of cases are 1+4+6+6+12+12+6+4+1= 52 cases
Hence the answer is 52.
Hope it was easy for you, too.