# How many strings of $8$ English letters are there (repetition allowed)?

a) at least one vowel

b) start with $x$ and at least one vowel

c) start and end with $x$ and at least one vowel

I can solve them easily by considering $total-no$ $vowel$. So,

a) $26^8 -21^8$

b) $26^7 -21^7$

c) $26^6 -21^6$

But, can I try

a) choose $1$ vowel out of $5$ in $5$ ways and put in one of $8$ places, and then the remaining places $26^7$, which gives $5\cdot 8 \cdot 26^7$? What should be the logic thinking this way?

• Classic case of overcounting. Arrangements with an a in the first position and an e in the second position are counted more than once. Commented Jun 8, 2015 at 20:59
• @AndréNicolas: Can you suggest me some other way apart from total-none ? Commented Jun 8, 2015 at 21:03
• I can suggest terrible ways: Exactly one position has a vowel, $8\cdot 5\cdot 21^7$. Exactly $2$ positions have a vowel: $\binom{8}{2}5^221^6$. Exactly $3$ positions have a vowel, $\binom{8}{3}5^3 21^5$. And so on. Add up. Or else use inclusion/exclusion to deal with the overcount in the method you suggested. Tricky. Commented Jun 8, 2015 at 21:09
• @AndréNicolas: Thanks for the solution, It is generalised for any number of cases. Commented Jun 9, 2015 at 7:25

An alternative to Joffan's solution is to count up all the ways there could be exactly $k$ vowels (as suggested by André Nicolas). We then get

$$N = \sum_{k=1}^8 \binom{8}{k} 5^k 21^{8-k}$$

All the methods yield $N = 171004205215$, confirming the expression $26^8-21^8$ you originally derived.

• Thanks, Can you tell me If atleast $3$ vowels are to be used ? Commented Jun 8, 2015 at 21:41
• Then it would be $\sum_{k=3}^8 \binom{8}{k} 5^k 21^{8-k}$. Commented Jun 8, 2015 at 21:46
• Thanks that would be helpful. Commented Jun 8, 2015 at 21:51

Your calculation has a lot of overcounting, whenever there is more than one vowel present.

If you really want to avoid (or, say, cross-check) the "negative space" method of simply excluding options with no vowels, you could perhaps sum through the possibilities of where the first vowel is:

• Vowel in first place: $5\cdot 26^7$ ways
• First vowel in second place: $21\cdot 5\cdot 26^6$ ways
• First vowel in third place: $21^2\cdot 5\cdot 26^5$ ways
• etc.

Total

$$5\cdot 26^7 + 21\cdot 5\cdot 26^6 + 21^2\cdot 5\cdot 26^5 + 21^3\cdot 5\cdot 26^4 \\ \quad\quad+ 21^4\cdot 5\cdot 26^3 + 21^5\cdot 5\cdot 26^2 + 21^6\cdot 5\cdot 26 + 21^7\cdot 5 \\ = 5 \sum_{k=0}^7 21^k \,26^{7-k}$$

Not easy.

Another method is to calculate options for an exact number of vowels. This can be calculated by setting the pattern in one step, eg for three vowels:

$$BBABBAAB$$

which is ${8 \choose 3}$, then multiplying by the options for consonants and vowels respectively, so

$${8 \choose 3}5^3\,21^5$$

for exactly three vowels. Then add up all options of interest (or, if simpler, add up the non-qualifying options and subtract from total).

• What if the question had stated atleast $2$ vowels or $3$ vowels or $4$ vowels etc etc. For atleast $2$ vowels, we could easily do $total-{none-one vowel}$. But if its $3$ vowels or $4$ vowels , then ? Do we have some nice and good appraoch ? Commented Jun 8, 2015 at 21:22
• answer extended Commented Jun 9, 2015 at 0:09