Answer to Question #342155 in Discrete Mathematics for syahidah

Question

a) A three digit number is to be formed using the digits 1, 2, 3, 4, 5, 6 and no repetition is

allowed.

i) How many numbers can be formed if the leading digit is 4?

ii) How many numbers can be formed if the number is more than 250?

iii) How many odd numbers can be formed between 200 and 400?

(8 marks)

b) Consider a bookshelf contains 28 books in different genre. 14 books are in education, 9

books in business and 5 books in motivation. A student would like to take 15 books. Find the number of ways if:

I) there is no restriction

ii) the choice must consist of 8 books in education, 5 books in business and 2 books in

motivation genre.

iii) The choice must consist of at least 9 books in education and exactly 5 books in

motivation genre.

(7 marks)

Answer 1

1.1 if the leading digit is 4 and no repetition is, we should choose 2 (r) digits from 5 (n). We have $\frac{n!}{(n-r)!}=\frac{5!}{(5-2)!}=20$ ways for it.

1.2. Situation 1

First number can be 3 -6. So we can choose it (1 digit from 4) in 4 ways.

Second digit can be 1-6, but it cannot repeat the first one, so we can choose it in 5 ways (1 from 5 digits).

Third digit can be 1-6, but it cannot repeat the first and the second digit, so we can choose it in 4 ways (1 digit from 4)

We can form 4x5x4=80 numbers.

Situation 2

First number is 2. Then we can choose first number in 1 way, then second number is 5-6 and we can choose it in 2 ways, and third digit can be 1-6, so we can choose it in 5 ways. So we can form 1x2x5=10 numbers.

Together we can form 80+10=90 numbers

1.3 First number can be 2 -4. So we can choose it (1 digit from 3) in 3 ways.

Second digit can be 1-6, but it cannot repeat the first one, so we can choose it in 5 ways (1 from 5 digits).

Third digit can be 1-6, but it cannot repeat first and second digit, so we can choose it in 4 ways (1 digit from 4)

We can form 3x5x4=60 numbers

2.1 $C^{15}_{28}=\frac{n!}{(n-r)!r!}=\frac{28!}{(28-15)!15!}=37442160$

2.2$C^8_{14}C^5_9C^2_5=\frac{14!}{8!6!}\frac{9!}{5!4!}\frac{5!}{2!3!}=3003x126x10=3783780$

2.3

We can take 9 books in education and 1 book in business or 10 in education and 0 books in business.

First situation (9 books in education and 1 in business)

$C^9_{14}C^1_9C^5_5=\frac{14!}{9!5!}\frac{9!}{8!1!}\frac{5!}{5!1!}=2002x9x1=18018$

Second situation (10 books in education and 0 in business)

$C^{10}_{14}C^0_9C^5_5=\frac{14!}{10!4!}\frac{9!}{9!0!}\frac{5!}{5!1!}=1001x1x1=1001$

So we have 18018+1001=19019 ways to choose books.