Answer to Question #343299 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?

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.

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

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

The 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

The first number is 2. Then we can choose the first number in 1 way, then the second number is 5-6 and we can choose it in 2 ways, and the 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 The first number can be 2 -4. So we can choose it (1 digit from 3) in 3 ways.

The 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).

The 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 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.

The first situation (9 books in education and 1 in business)

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

The 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.