Question

1. Assume that a word consists of 5 lower-case letters of the alphabet. Assume that repeated letters are NOT allowed to use in the word. How many words end with “t”? (4 points)
2. Assume that a word consists of 5 lower-case letters of the alphabet. If the word needs to contain letter “o”, how many such strings are there? (4 points)

Answer 1

( a )

Number of words end with “t” of lenth 5 = 25 * 24 * 23 * 22 = 303600

( b )

Number of words contain “o” of lenth 5 = 5 * 25 * 24 * 23 * 22 = 1518000

Explanation :

(a) Ending with it MI 25 ways 23 Ways ways I way es pays en ways ho says no says I way Number of wood of length 5 ending . with it when crepeation not allowed - 25 x 24*23* 22 XI 303,600 winds 'o' in werd (with No repeation) (b) Containity Case o ! letter when start with 23 Iway asways za vaje so says 22 ways Number of word z 25A 24 X 23 X22 Case 2! 'When second position letter o present at way I way ways

word 25 x 24x23722 No. of S I Case 3 when letter at thind position ' ť present 2.5 14оу оу 24oy погу ways No. of wind - 25X24 X23*22 o present at fourth position Caset letter 2tways ways way 22 ways os ways No. of word . = 25724823722 Case Sin when letter o presut at end zoways zaways 22 May 2lways I ways No. of word = 25*24x23x22 number of word=5*25*24 X23x22 -1518000 - Total