Puzzle - 123456789 = 100 with three operations? Given the sequence 123456789: You can insert three operations ($+$,$-$,$\\times$,$ $) into this sequence to make the equation = 100 My question is: is there a way to solve this without brute forc
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Combinations: 10 people divided in to two groups, one of 6 and one of 4? $$ \binom {10} {6} = \frac {10!} {6!\,4!} = \frac {10\cdot9\cdot8\cdot7\cdot\overbrace {6\cdot5\cdot4\cdot3\cdot2\cdot1}} {4\cdot3\cdot2\cdot1\cdot\underbrace {6\cdot5\cdot4\cdot3\cdot2\cdot1}} = \frac {10\cdot9\cdot8\cdot7} {4\cdot3\cdot2\cdot1} $$ Since $8$ cancels $4\cdot2$ and $\dfrac 9 3=3$, this reduces to $10\cdot3\cdot7$ (If you wanted two groups of $5$ then you'd do a similar thing