Frankenstein's answer is fast, elegant, and definitely the method you should use on the test. Since the vowels must be separate, they must have a consonant (or nothing) to the right and left. since there are only 5 consonants in each word, that leaves only 6 possible slots for the vowels in both problems (4 slots in between the consonants, and the two ends).
However, if you have difficulty understanding that method, here is another way to look at the problems:
1.) Another approach to problem 1 is to take the total number of permutations (ways you can arrange the letters in "numbers") and then subtract out the possibilities where the two vowels are together, to get the total arrangements where they are separate.
The total ways you can arrange the letters "NUMBER" is 5040. This is because in your 7-slot result, 7 letters could go in slot one, 6 letters could go in slot two, 5 in slot three, 4 in slot four, 3 in slot five, 2 in slot six, and 1 in the last slot. Multiplying, you get the permutations:
7 *
6 *
5 *
4 *
3 *
2 *
1 = 5040
Now how many of those permutations involve the two vowels being next to each other? To be next to each other, the two vowels could be in slots 1 and 2 together, slots 2 and 3, slots 3 and 4, slots 4 and 5, slots 5 and 6, or slots 6 and 7. That's 6 arrangements where the two vowels are next to each other. But it is actually 12, since for each arrangement the vowels could be in the order UE or EU.
To clarify, the 12 situations where the vowels are next to each other are:
E U _ _ _ _ _ ; U E _ _ _ _ _ ; _ E U _ _ _ _ ; _ U E _ _ _ _ ; _ _ E U _ _ _ ; _ _ U E _ _ _ ;
_ _ _ E U _ _ ; _ _ _ U E _ _ ; _ _ _ _ E U _ ; _ _ _ _ U E _; _ _ _ _ _ E U ; _ _ _ _ _ U E;
But for each of these 12 arrangements, there are actually many permutations, since the other letters can be arranged in different ways in the empty slots. Specifically, since there will always be 5 empty slots, the remaining letters can be arranged in 5 * 4 * 3 * 2 * 1 = 120 ways for each of the 12 arrangements where the vowels are adjacent. Thus, the total permutations where the vowels are adjacent would be 12 * 120 = 1,440.
So if you subtract the permutations where the vowels appear together (1,440) from the total permutations of the letters in "numbers" (5040), you get the permutations where they are separated: 3600.
2.) Following the same logic in problem #2, there are 36 different ways that at least two of the three vowels can be touching:
XXX_ _ _ _ _ ; XX_ X _ _ _ _ ; XX _ _ X _ _ _; XX _ _ _ X _ _; XX _ _ _ _ X _; XX _ _ _ _ _ X ;
_ XXX _ _ _ _; _ XX _ X _ _ _; _ XX _ _ X _ _; _ XX _ _ _ X _; _ XX _ _ _ _ X;
_ _ XXX _ _ _; _ _ XX _ X _ _; _ _ XX _ _ X _; _ _ XX _ _ _ X; X _ XX _ _ _ _;
_ _ _ XXX _ _; _ _ _ XX _ X _; _ _ _ XX _ _ X; X _ _ XX _ _ _; _ X _ XX _ _ _;
_ _ _ _ XXX _; _ _ _ _ XX _ X; X _ _ _ XX _ _; _ X _ _ XX _ _; _ _ X _ XX _ _;
_ _ _ _ _ XXX; X _ _ _ _ XX _; _ X _ _ _ XX _; _ _ X _ _ XX _; _ _ _ X _ XX _;
X_ _ _ _ _ XX; _ X _ _ _ _ XX; _ _ X _ _ _ XX; _ _ _ X _ _ XX; _ _ _ _ X _ XX;
Within those 36 arrangements the three vowels can be arranged 3!=6 different ways, so there are really 36 * 6 = 216 different arrangements where the vowels are not separate. In each of those different arrangements the remaining 5 letters can be arranged in 5! = 120 different ways so the total permutations where the vowels are NOT separate is 216 * 120 = 25,920. If you subtract the permutations where the vowels are NOT separate (25,920) from the total permutations of the 8 letters (40,320) you get the number of permutations where the vowels are separate: 14,400.
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