2

Good day

My question is as follows

Suppose the alphabet consists of ${A,B,C,D,E,F}$

(1) How many $4$-letter strings contain the word $“ACE”$?

(2) How many $4$-letter strings don’t have $“F”$ in the first position and $“E”$ in the last position?

(3) How many $5$-letter strings contain the word $“CAB”$?

(4) How many $4$-letter strings either begin with $“C”$ or end with two vowels?

So far, I answered that:

a) $6\times 2=12$ $ACE$ would appear either at one end or the other. Thus, one letter is optional and has $2$ positions.

b) $5\times 6\times 6\times 6\times 6\times 5=32400$

c) $6\times 6\times 4=144$

d) Begin with $C$ is $6\times 6\times 6=216$; end with two vowels $6\times 6\times 6\times 2\times 2=864$.

$6\times 2\times 2$ to get those not beginning with $C$ having two vowels at the end. $144$.

$6\times 4\times 4$ for those begininning with $C$, not having two vowels at the end. $96$.

Would this be the correct approach?

Thank you! -A

Jyotishraj Thoudam
  • 721
Dan Marchisella
  • 61
  • 1
    (2) is worded quite ambiguously. My reading would have been "(not (f at first)) and (e at last)". My second reading would have been "not (f at first and e at last")". However I don't manage to read from it the interpretation you apparently used, "(not (f at first)) and (not (e at last))". – celtschk Aug 08 '16 at 17:40

1 Answers1

0

Questions (1) and (2) look correct.

For question (3), there are three places the "cab" can be, and six choices for the other two letters.

For question (4), there are indeed $216$ four-letter words that begin with "c." For words that end in two vowels, it's $6(6)(2)(2) = 144$. (You have an extra $6$ in there.) Finally, subtract out those words that satisfy both to avoid double-counting: $(1)(6)(2)(2) = 24$. So $216 + 144 - 24 = 336$ words.

John
  • 26,319
  • Thus: b) C(5,1) * C(6,1)C(6,1) C(5,1) = 5325=900.

    c) 3 places "cab" could be, thus 6 options for the other letters. Thus, C(6,1)C(6,1)3=108.

    d) begin with C - 666=216; end with two vowels 662*2=144..

    Double Counted=(1)(6)(2)(2)=24.

    Thus, 216+144-24=336.

     – Dan Marchisella Aug 08 '16 at 18:59
  • Why do we include those double counted on 4? – Dan Marchisella Aug 08 '16 at 19:00
  • They were double counted because they were included in both the $216$ and the $144$. So by subtracting out one of the sets with the $24$, the double-counting problem is fixed. – John Aug 08 '16 at 19:11
  • perfect! thank you very very much! – Dan Marchisella Aug 08 '16 at 19:12