Permutations and Combinations

The reason that this is true is because we can simply fix 1 person to be at the top and there are (n − 1)! ways to arrange the other people. This accounts for rotations since rotating an arrangement will result in someone else on top.

We divide by 2 for reflections because of symmetry on both the left and right sides of the person chosen to be at the top.

The first digit has 5 choices (1, 3, 5, 7, and 9). The second digit, since it's an even number, also has 5 choices (0, 2, 4, 6, and 8). However, the third digit only has 4 choices, since we already selected an even number previously, and we only have 4 choices left. Same goes for the last digit, where it has 3 choices. We take all of these possible choices and multiply them together to get 5*5*4*3=300.

Without the condition, the answer would just be (6 − 1)! = 5! = 120 as we can fix 1 of the 6 people at the top and permute the remaining 5 people. However, that isn't the case, as we have constraints. Since there are only 6 people, there has to be one person seperating two of the enemies to get an alternating pattern, like X O X O X O. We can fix someone at the top, like Gauss (arbitrarily). We can now order the remaining 5 people. The location of the enemies is fixed, so there's 3! ways to order them. Now we just need to permute the remaining 2 physicists, and there's 2! ways. The answer is 3!2! which gives us (3*2*1) * (2*1) = 6*2 = 12.

This is like question 5, but harder. The first thing we should take note of is that the third digit should be a nonzero multiple of seven. This means that its only choice is to be 7, so the third digit is locked in as 7 and we cannot use it elsewhere, as all digits have to be distinct. Let's tackle the first 2 digits first. The question says that the first 2 digits must be 6 or more, so for the first digit, we have 6, 8, and 9 as our choices (remember that we can't use 7), giving us 3 possible choices for the first digit. For the second digit, we have 2 choices left, as one of the 3 choices was taken by the first digit. So far we have 3, 2, and 1 combinations. For the last 2 digits, they have to be 5 or less, so we can have 5, 4, 3, 2, 1, and 0, giving us 6 choices. The last digit has 5 choices, as one was taken before. Now for the fourth digit, we just need to see all the numbers that we have left. Let's assume that we took 6, 8, 7, _, 5, and 4 as our digits. The 4th digit has 0, 1, 2, 3, and 9 as its choices, as all the other choices are taken, giving us 5 combinations. When we multiply all of these combinations together, 3 * 2 * 5 * 6 * 5, we get 30 * 30, and we get 900 as our answer.

For the first item, we have 3 choices. Pencil, Eraser, and Marker. Let's assume he puts in an eraser (arbitrarily). For the second item, he has 2 choices. Marker or Pencil. Again, let's assume he puts in a marker (arbitrarily). For the third item, he only has one choice, pencil. For the fouth, fifth, and sixth, it's still the same. 3 choices, then 2, then one. For the last item, it has to be a pencil, since there's none of anything else left. However, we're not done. You'd assume that it's just all the combinations together (3 * 2 * 1 * 3 * 2 * 1 * 1 = 36), but you need to account for thte fact that the pencils are distinct, the erasers, are distinct, and the markers are distinct. For this, we need to multiply the original answer (36) by 3! (Pencils), 2! (Erasers), and 2! (Markers) to get 36 * 3!2!2! = 36 * 24 = 864.

where n is the number of letters and d1, d2, d3, . . . are the number of times each of the letters that occur more than 1 time appear in the word.

For this word rearrangement problem, let's treat the COMP component as a single "block", like [COMP]UTER. Now the total amount of ways to rearrange this is 5!, since the [COMP] block is treated as a single letter. BUT, you can still rearrange the letters in [COMP], as they don't necessarily have to stay in the same order, so that's another 4! Multiplying 5! and 4! together gives you our answer of 2880.

Think of it like Column A as AAA, Column B as BBB, and Column C as CC. Since the targets aren't distinct, then we can use the word rearrangement formula. AAABBBCC. The total ways to rearrange it is 8!/3!3!2!, giving us 560.

A, M, O, S, and U are the alphebetical order of these letters. We need to find the order, since U is last here. A____ has 24 combinations, and so does M____, O____, and U____, giving us 96 "words" before we even reach U____. The thing is, S is also the second-to-last, so: UA___ has 6 combinatinos, UM___ has 6, and UO___ has 6. When we get to US___, the first "word" we have in alphebetical order is just USAMO, so 24+24+24+24+6+6+6 = 96+18 = 114 is the answer, right? No. We need to add one, since the list starts on one, not zero, giving us our answer of 115.

Permutations' order matters, combinations' don't.

This can be said as "n choose k".

This is true because we can see choosing k objects on the left hand side is equivalent to the n − k objects that will not be selected on the right hand side.

We must choose 2 managers from the 6 experienced applicants, so it's 6 choose 2, giving us 6!/2!4!. This simplifies into 6*5/2 = 15. Now, we need to select the other employees. There's 11 unhired people left, so we just do 11 choose 4, giving us 330.15 * 330 = 4950, which is our answer.

Since the energy bars are different, we'll have to distribute them with the combinations formula. Let's do Alice first. She needs 3 bars, and we have 12. We can just do 12 choose 3, which gives us 220. There's 9 bars left, so now let's do Betty. She needs 4 bars, so we'll do 9 choose 4, badabing, badaboom, we get 126 combinations for Betty. For Chase, there's 5 bars left, and he needs 5, so 5 choose 5 is just one. From all this, we multiply 220, 126, and 1, to get 27,720 as our answer.

First off, every row must have exactly one ball in it, as if one row was empty, a ball would be placed in an already occupied row, so that wouldn't work. Now how do we figure out the columns?... Well, we can choose 4 out of the 6 to have a ball in them, so it's 6 choose 4, giving us 15. Let's say we have our 4 rows selected, how will we place our balls in there now (pause)? Well, for our first row, we have 4 choices of columns for it to be placed in, giving us 4 choices. For the second ball, one column is taken up, so we have 3 choices. It repeats until we get to the last ball, so the ball positions are 4!. Because of this, our answer is just the number of ways to place the balls and the number of column combinations we have, so it's 24 × 15, giving us 360 as our answer.

We can use the word rearrangement formula for 8!/4!4! to get 70. We'll now number the ways to put the coins face up or face down. If a coin's face down, then the next coin can be anything and their faces will never meet. If a coin's face is up, then the next coin must face up as well. The moment a coin becomes face up, all the other next coins must also be face up as well. There's 8 ways to do this. If none of the coins are face up, there's one orientation. In total, we have 9 (8+1) orientations, and 9*70 is 630.

We have 2 choices for each element in the set: whether to include or not include the element in our subset, and since there are n elements in the set, the total number of subsets is 2 × 2 × 2 × · · · × 2 (n times).

First, let's choose which 2 will be the intersection. Five choose two gets us 10 combinations. We must then divide the remaining 3 elements among two subsets. Each element has two choices; whether to be in subset A or subset B, so we have 2^3=8 choices for the subsets. 8*10 = 80. Because the order doesn't matter, we have to divide by two for symmetry, as we can just flip them.