Hello, cuteisa89!

Of 10 candidates, the first 6 are being interviewed in the morning, and the last 4 in the afternoon.

In how many ways can the interviewer arrange a schedule?

I will assume that theorderof the candidates is important.

For the morning interviews, there are P(10,6) = 10·9·8·7·6·5 = 151,200 arrangements.

For the afternoon interviews, there are P(4,4) = 4·3·2·1 = 24 arrangements.

There are: .151,200 x 24 .= .3,628,800 possible schedules.

How many seven-digit phone numbers can be formed if the first digit cannot be 0

and (a) repetition of digits is not permitted? . (b) repetition of digits is permitted?

(a) The first digit can be any of 9 digits (1 to 9).

The second can be any of the remaining 9 digits.

The third can be any of the remaining 8 digits.

The fourth can be any of the remaining 7 digits.

The fifth can be any of the remaining 6 digits.

The sixth can be any of the remaining 5 digits.

The seventh can be any of the remaining 4 digits.

Answer: .9·9·8·7·6·5·4 .= .544,320

(b) The first digit can be any of 9 digits (1 to 9)

The second through the seventh can be any of the 10 digits.

Answer: .9·10·10·10·10·10·10 .= .9,000,000

Using the letters of the word YOUNG, how many different 5-letter combinations

are possible if: (a) the first letter must be Y?

(b) the vowels and consonants alternate, beginning with a consonant?

(Y is a consonant here andY does not have to be first) What does this mean?

(a) The first letter is Y.

Then the letters {O,U,N,G} can be arranged in P(4,4) ways.

Answer: .4·3·2·1 .= .24 combinations.

(b) The letters must be: consonant-vowel-consonant-vowel-consonant.

The three consonants can be placed in 3! = 6 ways.

The two vowels can be be placed in 2! = 2 ways.

. . There are: .6 x 2 .= .12 combinations.