Hey there!

For the first question, you use that identity in conjunction with a bit of algebra to work your way around.

Second question is reasonably simple. Since you can put each potato in any of the five boxes, and it doesn't matter how many each gets, you have 5 boxes to put in the first potato, 5 boxes to put in the second potato, 5 boxes to put in the third potato. So the answer is just 5

^{3} or 125.

Third question: You have three ways of arranging them: 2 rows of 2, 1 row of 3 and 1 of 1, 1 row of 4. With the 2 rows of two, effectively what you need to do is change the diagonals ie. where the Y's or X's are. You need to change what is diagonally adjacent to a fixed shirt, since basically most of the other combos are just rotations. for this part you get 3x2!x2! (try drawing them out, it'll make sense. theres only 12 anyway). Then, for the row of 3 and the row of 1 it's just 3! (for the row of 3) x 4 (for the different shirts in the row of 1) and the last is 4!. Add them all up and you should get 60.

YX

XY

Fourth question: not too sure, gonna come back to this one

Fifth question: Since you have 3 onions and 2 zucchini, they have to go in the order OZOZO. With this set order, it's easy to see that the answer will be 3!x2!, or 12.

Sixth Question:

a) Let the last number be 5. Then the first digit must be non-zero ie.1, 2, 3, 4. The second digit must be the three not used and zero, and the third must be the three not used. So with 5 as the last digit, there are 4x4x3x1 combos. With zero as the last digit, there are 5 possibilities for the first digit, four, then three for the respective digits. This becomes 5x4x3x1. Add those together, you get 48+60=108.

b) Basically everything with less than 5 digits

4 digits -- 5x5x4x3 --> the first number thing is for the non-zero, seen above in a)

3 digits -- 5x5x4

2 digits -- 5x5

1 digit -- 6 (we can include zero)

add them up, you should get 431.

c) Set the last digit to be a non zero even number (2 or 4). Then the first digit must be non-zero, and there are four options, then including zero with the other three leftovers, we get 4x4 x2(for the two different numbers). Then if we stick zero at the end, there is an additional option for the first digit, making it 5x4x1. Add them up and you should get 52.

Seventh question -- the intuition behind this is that the first number is the same as the last, second same as second last and so on. For the five digit number we have ABCBA. There are 9 options for A (A is a non-zero digit) while for B and C we have 10 options. 9x10x10=900. The same thing applies for the six digit number since you're effectively making it ABCCBA.

Hope this helps