Raising Squares Part 1

We know for sure that the minimum resultant that we can get is 1. So, keeping the number 1 in the huge box, the other three numbers 2,3,4 in the smaller boxes can be changed. So, 3! possible arrangements can be obtained where the resultant number is 1

It is for sure that the minimum number will be 1. So in 1st box .it'll be 1. And in rest 3 boxes 2,3,4can be arranged in 3! Ways that is six .so and. Is SIX

The min. resultant is obviously one, so we need to calculate the number of ways we can get it (by using simple combinatorics):
_ * _ * _
3 * 2 * 1
=> n=6, where n is the number of ways to get 1.

4-1=3!=6 that’s wassup

If the lowest base is $1$ , then it doesn't matter what the following powers are in the exponent tower; the result will still be 1. Therefore, $S=1$ . This leaves $2, 3,$ and $4$ . We may only choose one of these three digits for the three remaining positions, and once a digit is chosen, it may not be chosen again. As such, this is a straightforward combinatorics problem: 3 choices $\rightarrow$ 2 choices $\rightarrow$ 1 choice: $3 \cdot 2 \cdot 1 = 6$

In other words, we have 3 possible digits and 3 slots to fill. When the number of slots to fill $k$ is the same as the number of digits available (without replacement) $n$ , the number of possible combinations is simply $n!$ .

We sure that the minimum value is 1 because it's contains 1.Odd 1 out & we have $(4-1)!=3!=\boxed{\large{6}}$