Thread: Help Please Combinations

Hi, I don't know how to solve these problems. Can someone help me?

1. 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?
2. How many seven-digit phone numbers can be formed if the first digit cannot be 0 and repetition of digits is not permitted? if repetition of digits is permitted?
3. Using the letters of the word YOUNG, tell how many different 5-letter combinations are possible if: the first letter must be Y? the vowels and consonants alternate, beginning with a consonant (Y is a consonant here and Y does not have to be first)?

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 the order of 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 and Y 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.