Q.

Find the numer of permutations of five different objects taken three at a time.

➤

Solution:
   The number of permutations of five different objects taken three at a time is \(5P3 = \frac{5!}{(5-3)!} = 5 \cdot 4 \cdot 3 = 60\).
 

Q.

If three persons enter a bus in which there are ten vacant seats, find in how many ways they can sit.

➤

Solution:

    The number of ways in which three persons can sit in ten vacant seats is \(10P3 = \frac{10!}{(10-3)!} = 10 \cdot 9 \cdot 8 = 720\).

Q.

1. How many plates of vehicles consisting of 4 different digits can be made out of the integers 4, 5, 6, 7, 8, 9? How many of these numbers are divisible by 2?
2. How many numbers of 4 different digits can be formed from the digits 2, 3, 4, 5, 6, 7? How many of these numbers are
   1. divisible by 5
   2. not divisible by5.
3. How many 5 digit odd numbers can be formed using the digits 3, 4, 5, 6, 7, 8 and 9. If 
   1. repetition of digits is not allowed
   2. repetition of digits is allowed

➤

Solution:

1. The number of plates of vehicles consisting of 4 different digits can be made out of the integers 4, 5, 6, 7, 8, 9 is \[6P4 = \frac{6!}{(6-4)!} = 6 \cdot 5 \cdot 4 \cdot 3 = 360\]
   The number of these numbers that are divisible by 2 is \[5P3 = \frac{5!}{(5-3)!} = 5 \cdot 4 \cdot 3 = 60\]
2. The number of numbers of 4 different digits that can be formed from the digits 2, 3, 4, 5, 6, 7 is \[6P4 = \frac{6!}{(6-4)!} = 6 \cdot 5 \cdot 4 \cdot 3 = 360\]
3.  
   1. The number of 5 digit odd numbers that can be formed using the digits 3, 4, 5, 6, 7, 8 and 9 without repetition is \[6P5 = \frac{6!}{(6-5)!} = 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 = 720\]
   2. The number of 5 digit odd numbers that can be formed using the digits 3, 4, 5, 6, 7, 8 and 9 with repetition is \[7 \cdot 6 \cdot 5 \cdot 4 \cdot 5 = 8400\]

Q.

In how many ways can four boys and three girls be seated in a row containing seven seats

1. if they may sit anywhere
2. if the boys and girls must alternate
3. if all three girls are together
4. if girls are to occupy odd seats

➤

1. The number of ways four boys and three girls can be seated in a row containing seven seats if they may sit anywhere is \(7!\).
      \[
          7! = 5040
      \]
2. If the boys and girls must alternate, we can start with a boy only so as to accomodate the three girls between the 4 boys. There is only one possible patterns: BGBGBGB. For this pattern, we can arrange the boys in \(4!\) ways and the girls in \(3!\) ways. Therefore, the total number of ways is:
      \[
          4! \times 3! = 24 \times 6 = 144
      \]
3. If all three girls are together, we can treat the three girls as a single unit. So, we have a total of 5 units to arrange in \(5!\) ways. The three girls within their unit can be arranged in \(3!\) ways. Therefore, the total number of ways is:
      \[
          5! \times 3! = 120 \times 6 = 720
      \]
4. If the girls are to occupy odd seats, it means that the girls can't occupy the even seats, but the boys may occupy odd or even seats. The odd seats are 1, 3, 5, and 7. The question also implies that only boys can occupy the even seats. The even seats are 2, 4, and 6 i.e. 3 seats. So, the 4 boys can be arranged in these 3 seats in \[P(4, 3) = \frac{4!}{(4-3)!} = 4 \cdot 3 \cdot 2 = 24\] ways. The 3 girls and 1 boy can be arranged in the 4 odd seats in \[P(4, 4) = \frac{4!}{(4-4)!} = 4 \cdot 3 \cdot 2 \cdot 1 = 24\] ways. Therefore, the total number of ways is:
      \[
          24 \times 24 = 576
      \]

Q.

In how many ways can eight people be seated in a row of eight seats so that two particular persons are

1. always together
2. never together

➤

1. If two particular persons are always together, we can treat the two persons as a single unit. So, we have a total of 7 units to arrange in \(7!\) ways. The two persons within their unit can be arranged in \(2!\) ways. Therefore, the total number of ways is:
      \[
          7! \times 2! = 5040 \times 2 = 10080
      \]
2. The total number of arrangements of 8 people in 8 seats is \(8!\). The number of ways the two particular persons are never together is the total number of arrangements minus the number of ways the two particular persons are always together. Therefore, the total number of ways is:
      \[
          8! - 10080 = 40320 - 10080 = 30240
      \]

Q.

Six different books are arranged on a shelf. Find the number of different ways in which the two particular books are

1. always together
2. not together

➤

1. If the two particular books are always together, we can treat the two books as a single unit. So, we have a total of 5 units to arrange in \(5!\) ways. The two books within their unit can be arranged in \(2!\) ways. Therefore, the total number of ways is:
      \[
          5! \times 2! = 120 \times 2 = 240
      \]
2. The total number of arrangements of 6 books is \(6!\). The number of ways the two particular books are not together is the total number of arrangements minus the number of ways the two particular books are always together. Therefore, the total number of ways is:
      \[
          6! - 240 = 720 - 240 = 480
      \]

Q.

In how many ways can four red beads, five white beads and three blue beads be arranged in a row?

➤

The number of ways four red beads, five white beads and three blue beads can be arranged in a row is
   \[ \frac{n!}{p!\cdot q! \cdot r!}\]
   where \(n = 4 + 5 + 3 = 12\), \(p = 4\), \(q = 5\), and \(r = 3\). Therefore, the total number of ways is:
   \[
       \frac{12!}{4! \cdot 5! \cdot 3!} = 27720
   \]

Q.

In how many ways can the letters of the following words be arranged?

1. ELEMENT
2. NOTATION
3. MATHEMATICS
4. MISSISSIPI

➤

1. The number of ways the letters of the word ELEMENT can be arranged is
      \[
          \frac{7!}{3!} = 840
      \]
2. The number of ways the letters of the word NOTATION can be arranged is
      \[
          \frac{8!}{2! \cdot 2! \cdot 2!} = 5040
      \]
3. The number of ways the letters of the word MATHEMATICS can be arranged is
      \[
          \frac{11!}{2! \cdot 2! \cdot 2!} = 4989600
      \]
4. The number of ways the letters of the word MISSISSIPI can be arranged is
      \[
          \frac{11!}{4! \cdot 4! \cdot 2!} = 34650
      \]

Q.

How many numbers of 6 digits can be formed with the digits 2, 3, 2, 0, 3, 3?

➤

Since, 0 cant be placed at the first place, the number of numbers of 6 digits that can be formed with the digits 2, 3, 2, 0, 3, 3 is
   \[
       \frac{5 \cdot 5!}{2! \cdot 3!} = 50
   \]

Q.

In how many ways can 4 art students and 4 science students be arranged in a circular table if

1. they may sit anywhere
2. they sit alternately

➤

1. The number of ways 4 art students and 4 science students can be arranged in a circular table if they may sit anywhere is
      \[
          7! = 5040
      \]
2. If they sit alternately, we can start with an art student only so as to accomodate the 4 science students between the 4 art students. There is only one possible patterns: ASASASAS. For this pattern, we can arrange the art students in \(4!\) ways and the science students in \(4!\) ways. Therefore, the total number of ways is:
      \[
          3! \times 4! = 6 \times 24 = 144
      \]

Q.

In how many ways can eight people be seated in a round table if two people insist in sitting next to each other?

➤

The number of ways eight people can be seated in a round table if two people insist in sitting next to each other is
   \[
       6! \times 2! = 1440
   \]

Q.

In how many ways can seven different coloured beads be made into a bracelet?

➤

The number of ways seven different coloured beads can be made into a bracelet is
   \[
       6! = 720
   \]
   But, since the bracelet can be rotated, the number of ways is
   \[
       \frac{6!}{2} = 360
   \]

Q.

1. In how many ways can 4 letters be posted in six letter boxes?
2. How many even numbers of 3 digits can be formed when repetition of digits is allowed?
3. In how many ways can 3 prizes be distributed among 4 students so that each student may receive any number of prizes?

➤

1. The number of ways 4 letters can be posted in six letter boxes is
      \[
          6^4 = 1296
      \]
2. The number of even numbers of 3 digits that can be formed when repetition of digits is allowed is
      \[
          5 \cdot 10 \cdot 9 = 450
      \]
3. The number of ways 3 prizes can be distributed among 4 students so that each student may receive any number of prizes is
      \[
          4^3 = 64
      \]

Q.

In how many ways can the letters of the word "MONDAY" be arranged? How many of these arrangements do not begin with M? How many begin with M and do not end with Y?

➤

The number of ways the letters of the word "MONDAY" can be arranged is
   \[
       {6!}= 720
   \]

   The number of these arrangements that do not begin with M is
   \[
       5\cdot 5! = 600
   \]

   The number of these arrangements that begin with M and do not end with Y is
   \[
       1 \cdot 4 \cdot 4! = 96
   \]

Q.

Show that the number of ways in which the letters of the word

1. "COLLEGE" can be arranged so that the two E's always come together in 360.
2. "ARRANGE" can be arranged so that no tow R's come together is 900.

➤

1. The number of ways the letters of the word "COLLEGE" can be arranged so that the two E's always come together is
      \[
          \frac{6!}{2!} \cdot 2! = 360
      \]
2. The number of ways the letters of the word "ARRANGE" can be arranged so that no two R's come together is
      \[
          \frac{7!}{2!} - 2 \cdot \frac{6!}{2!} = 900
      \]

Q.

In how many ways can the letters of the word "COMPUTER" be arranged so that

1. all the vowels are always together?
2. the vowels may occupy only odd positions?
3. the relative positions of vowels and consonants are not changed?

➤

1. Total numbers of words in the word "COMPUTER" is 8.
      The number of vowels is 3 and the number of consonants is 5.
      The number of ways the letters of the word "COMPUTER" can be arranged so that all the vowels are always together is
      \[
          \frac{6!\cdot 3!} = 4320
      \]
2. The odd positions are 1, 3, 5, and 7, i.e, 4 positions. The number of ways the consonants can be arranged in these 4 positions is \[P(5, 4) = \frac{5!}{(5-4)!} = 5 \cdot 4 \cdot 3 \cdot 2 = 120\]. The vowels and 1 remaining consonant can be arranged in the remaining 4 positions in \[P(4, 4) = \frac{4!}{(4-4)!} = 4 \cdot 3 \cdot 2 \cdot 1 = 24\] ways. Therefore, the total number of ways is:
      \[
          120 \times 24 = 2880
      \]
3. The question implies that the vowels and consonants remain in their relative positions. The number of ways the vowels can be arranged is \(P(3, 3) = \frac{3!}{(3-3)!} = 3 \cdot 2 \cdot 1 = 6\). The number of ways the consonants can be arranged is \(P(5, 5) = \frac{5!}{(5-5)!} = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120\). Therefore, the total number of ways is:
      \[
          6 \times 120 = 720
      \]

Q.

Find the number of arrangements of the letters of the word "LAPTOP" so that

1. the vowels may never be separated;
2. all consonants may not be togeter;
3. they always begin with L and end with T
4. they do not beign with L but always end with T.

➤

The word "LAPTOP" has 6 letters, 2 vowels (A, O) and 4 consonants (L, P, T, P).

1. The number of ways the letters of the word "LAPTOP" can be arranged so that the vowels may never be separated is
      \[
          \frac{5! \cdot 2!}{2!} = 120
      \]
2. The number of ways the letters of the word "LAPTOP" can be arranged so that all consonants may not be together is
      \[
          \frac{6!}{2!}  - \frac{4! \cdot 3!}{2!} = 360 - 72 = 288
      \]
3. The number of ways the letters of the word "LAPTOP" can be arranged so that they always begin with L and end with T is
      \[
          \frac{1\cdot 1\cdot 4!}{2!} =   12
      \]
4. The number of ways the letters of the word "LAPTOP" can be arranged so that they do not begin with L but always end with T is
      \[
          \frac{1\cdot 4 \cdot 4!}{2!} = 48
      \]

Q.

How many different words can be formed with all the letters of the word "Internet" if

1. each word is to begin with vowel?
2. each word is to end with consonant?

➤

1. The word "Internet" has 8 letters, 3 vowels (I, e, e) and 5 consonants (n, t, r, n, t). The number of ways the letters of the word "Internet" can be arranged so that each word is to begin with a vowel is
      \[
          \frac{3 \cdot 7!}{2!\cdot 2! \cdot 2!} = 1890
      \]
2. The number of ways the letters of the word "Internet" can be arranged so that each word is to end with a consonant is
      \[
          \frac{5 \cdot 7!}{2!\cdot 2! \cdot 2!} = 3150
      \]