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If the alphabet consists of $\{a,b,c,d,e,f\}$, how many four letter strings either start with c or end with two vowels?

My reasoning was as follows:

Starts with c but does not have vowel in third position + starts with c but does not have a vowel in fourth position + does not start with c but has vowels in the last two positions $$= 1 \cdot 6 \cdot 4 \cdot 6 + 1 \cdot 6 \cdot 6 \cdot 4 + 5 \cdot 6 \cdot 2 \cdot 2 = 408$$

Is my reasoning correct?

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  • $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ Jun 5 '17 at 0:13
  • $\begingroup$ Your forgot starts with c and ends with two vowels 1.6.2.2 which is acceptable. $\endgroup$
    – fleablood
    Jun 5 '17 at 1:11
  • $\begingroup$ Thanks guys, after doing enough of these problems, I came up with the easiest method. For all strings that start with c but do not end with two vowels is C(1,1)*C(6,1)*((6*6)-4)=192. All strings that do not start with c but end with two vowels is C(5,1)*C(6,1)*C(2,1)*C(2,1)=120. 192+120=312 $\endgroup$ Jun 15 '17 at 0:57
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Method 1: The number of strings that start with c is $$1 \cdot 6 \cdot 6 \cdot 6 = 216$$ since there is only one way to fill the first position with c and there are six ways to fill each of the remaining positions with one of the six letters in the alphabet.

The number of strings that end with two vowels is $$6 \cdot 6 \cdot 2 \cdot 2 = 144$$ since the first two positions can be filled with any of the six letters of the alphabet and the last two positions can be filled with either of the two vowels.

If we add the above cases, we will have counted strings that start with c and end with two vowels twice. Since we only want to count them once, we must subtract such arrangements from the total. There are $$1 \cdot 6 \cdot 2 \cdot 2 = 24$$ such strings.

Hence, the number of strings that start with c or end with two vowels is $$216 + 144 - 24 = 336$$

If you instead meant that exactly one of the following is true: starts with c or ends with two vowels, then we must subtract the $24$ cases in which a string both starts with c and ends with two vowels from the above result to obtain $$216 + 144 - 2 \cdot 24 = 312$$

Method 2: Let's fix your count to address the issues @Dana found with your reasoning. We must consider disjoint cases:

Starts with c and ends with two consonants: The number of such strings is $$1 \cdot 6 \cdot 4 \cdot 4 = 96$$

Starts with c and has a vowel in the third position but not the fourth: The number of such strings is $$1 \cdot 6 \cdot 2 \cdot 4 = 48$$

Starts with c and has a vowel in the fourth position but not the third: The number of such strings is $$1 \cdot 6 \cdot 4 \cdot 2 = 48$$

Starts with a letter other than c and ends with vowels in the last two positions: The number of such strings is $$5 \cdot 6 \cdot 2 \cdot 2 = 120$$

Starts with c and ends with vowels in the last two positions: The number of such strings is $$1 \cdot 6 \cdot 2 \cdot 2 = 24$$

If or is used in the inclusive sense (as it usually is in mathematics), we obtain a total of $$96 + 48 + 48 + 120 + 24 = 336$$ strings that begin with c or end with two vowels.

If or is used in the exclusive sense, we obtain a total of $$96 + 48 + 48 + 120 = 312$$ strings that begin with c or end with two vowels, but not both.

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Easier to do inclusion-exclusion.

Starts with C + ends with two vowels - (starts with C and ends with two vowels)

$1*6*6*6 + 6*6*2*2 - 1*6*2*2 = 336$

You were okay but a double adding snuck in. You counted those that start with C and do not have vowels in either the 3rd or 4th position twice.

Because those things have a way of sneaking in and as if this got even the tiniest bit more complicated you'd have many many things to keep track of, I prefer the inclusion-exclusion method.

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