How many different five-letter words can be formed using the letters of the word STEVIN,
a) if only one vowel is to be used?
b) if both vowels are to be used and E is to precede I?
c) if both vowels must be used, they must be adjacent, and E is to precede I?
d) if either one vowel is to be used or E is to precede I?
You must be familiar with the usual conventions of permutations, P(i,j), the number of ways of ordering i items out of j distinct items, where
P(i,j)=j!/(j-i)!
For example, how many 2 lettered words can be made of the letters a,b and c:
P(2,3)=3!/(3-2)!=3!/1!=6
a)
You would be making 5 lettered words out of the 5 letters STEVN or STVIN.
The number of words:
P(5,5)+P(5,5)
b)
You'd be making words out of
STEVI, STEIN, SEVIN, TEVIN
for a total of
4P(5,5) words.
However, half of these words have the vowels in wrong order, so the answer should be 2P(5,5).
c)
Treat EI as one letter X, and choose 3 remaining letters to give
STVX, STNX, SVNX, TVNX
So you have 4P(4,4) words.
d)
The two cases are mutually exclusive, so we just have to add cases a) and c).
Check my work.
a) Well, if only one vowel is to be used, we have to choose between E and I. So the answer is 2!
b) If both vowels are to be used and E has to precede I, we can think of it like this: E is the aspiring leader and I is the follower. No room for debate! So we have fixed positions for E and I, leaving 3 remaining letters to arrange. The answer is therefore 3!
c) If both vowels have to be used, they must be adjacent, and E has to precede I, it's like a vowel power couple! E and I are inseparable. So we can consider them as one unit. This leaves us with S, T, V, and N. So the answer is 4!
d) If either one vowel is to be used or E is to precede I, it's a bit more flexible. We have two scenarios to consider. Firstly, if either E or I is used, then we have 5 choices in total. Secondly, if E has to precede I, we can think of it as one unit again, which leaves us with 4 remaining letters to arrange. So the answer is 5 + 4 = 9, combining the two scenarios!
Hope that puts a smile on your face!
a) To count the number of different five-letter words that can be formed using the letters of the word STEVIN, with only one vowel used, we need to consider the different cases:
Case 1: The vowel is E.
In this case, we have 1 vowel (E) and 4 consonants (S, T, V, N). The vowel can occupy any of the 5 positions in the word. The consonants can be arranged in the remaining 4 positions. So, the number of words is 5 * 4! = 120.
Case 2: The vowel is I.
Following the same logic as in Case 1, the number of words is also 120.
Therefore, the total number of different five-letter words with only one vowel used is 120 + 120 = 240.
b) To count the number of different five-letter words that can be formed using the letters of the word STEVIN, with both vowels used and E preceding I, we can consider the following cases:
Case 1: The word starts with EI.
In this case, we have 1 vowel (E) and 1 consonant (S) before the vowel I. The vowel E can occupy the first position, and the vowel I can occupy the second position. The consonants can be arranged in the remaining 3 positions. So, the number of words is 1 * 1 * 3! = 6.
Case 2: The word starts with I, followed by E.
In this case, we have 1 vowel (I) and 1 consonant (S) before the vowel E. The vowel I can occupy the first position, and the vowel E can occupy the second position. The consonants can be arranged in the remaining 3 positions. So, the number of words is 1 * 1 * 3! = 6.
Therefore, the total number of different five-letter words with both vowels used and E preceding I is 6 + 6 = 12.
c) To count the number of different five-letter words that can be formed using the letters of the word STEVIN, with both vowels adjacent, and E preceding I, we can consider the following case:
Case 1: The word starts with EI.
In this case, we have 1 vowel pair (EI) and 3 consonants (S, T, V) before the vowel I. The vowel pair EI can occupy the first two positions. The consonants can be arranged in the remaining 3 positions. So, the number of words is 1 * 3! = 6.
Therefore, the total number of different five-letter words with both vowels adjacent and E preceding I is 6.
d) To count the number of different five-letter words that can be formed using the letters of the word STEVIN, with either one vowel used or E preceding I, we can consider the following cases:
Case 1: Only one vowel is used.
Following the logic from part a), there are 240 possible words.
Case 2: E precedes I.
Following the logic from part b), there are 12 possible words.
Therefore, the total number of different five-letter words with either one vowel used or E preceding I is 240 + 12 = 252.
To find the number of different five-letter words that can be formed using the letters of the word STEVIN, we need to consider the given conditions in each case.
a) If only one vowel is to be used:
In the word STEVIN, there are three vowels (E, I, and I). We need to choose one vowel from these three and arrange the remaining four consonants (S, T, V, and N).
Number of ways = (3 vowels choose 1 vowel) * (4 consonants arrange) = 3 * 4! = 3 * 24 = 72.
b) If both vowels are to be used and E is to precede I:
In this case, we have two vowels (E and I) that must be used, and E should come before I. The remaining three consonants (S, T, and N) need to be arranged.
Number of ways = (2 vowels choose 2 vowels) * (3 consonants arrange) = 1 * 3! = 1 * 6 = 6.
c) If both vowels must be used, they must be adjacent, and E is to precede I:
Here, we have two vowels (E and I) that must be used, and they have to be adjacent with E preceding I. In this case, we can treat "EI" as a single entity. The remaining three consonants (S, T, and N) also need to be arranged.
Number of ways = (2 vowels choose 2 vowels) * (3 consonants arrange) = 1 * 3! = 1 * 6 = 6.
d) If either one vowel is to be used or E is to precede I:
In this case, we can calculate the total number of words formed by using "at least one vowel" and subtract the number of words formed by using both vowels (as required in case b) from it.
Total number of words using at least one vowel = Total number of words in the given word - Number of words using no vowels
Total number of words using at least one vowel = 6! - 4! (considering "E" and "I" as one consonant)
Number of words using either one vowel or E precedes I = Total number of words using at least one vowel - Number of words using both vowels as required in case b.
Number of words using either one vowel or E precedes I = (6! - 4!) - 6 = 720 - 24 - 6 = 690.
Therefore, the answers to the given cases are:
a) 72 different five-letter words.
b) 6 different five-letter words.
c) 6 different five-letter words.
d) 690 different five-letter words.