



 
Hi Evan, Let me look at going from 3 to 4. There are 3 × 2 × 1 = 6 permutations of the numbers 1, 2 and 3. (Permutation means that order is important so for example 123 is a different permutation than 132.) There are so few permutations of 1, 2 and 3 that I can list them
I can extend each one of them to a permutation of 1, 2, 3 and 4 by inserting the digit 4
For example 213 can be extended to
Thus there are 4 times as many permutations of 1, 2, 3 and 4 as there are permutations of 1, 2 and 3. Hence there are 4 × 3 × 2 × 1 = 24 permutations of 1, 2, 3 and 4. Now I can extend each of these to a permutation of 1, 2, 3, 4 and 5 by inserting the digit 5
Hence there are 5 × 4 × 3 × 2 × 1 = 120 permutations of 1, 2, 3, 4 and 5. And so on,  


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