Shortcut Methods
JEE Mains/Advanced
1. Find the number of ways in which 5 balls of different colours can be arranged in a row.
Shortcut Method: Use the permutation formula: $$nPr = \frac{n!}{(nr)!}$$ where n is the total number of items and r is the number of items to be selected. In this case, n = 5 and r = 5. $$5P5 = \frac{5!}{(55)!} = \frac{5!}{0!} = 5! = 120$$ Therefore, there are 120 ways in which 5 balls of different colours can be arranged in a row.
2. Find the number of ways in which 10 students can be seated in a row if 2 particular students always sit together.
Shortcut Method: Treat the 2 students who always sit together as one unit. So, we have 9 students and 1 unit to arrange. $$9P10 = \frac{9!}{(910)!} = \frac{9!}{1!} = 0$$ Therefore, there is no way to arrange 10 students in a row if 2 particular students always sit together.
3. Find the number of ways in which 6 letters of the word ‘MATHEMATICS’ can be arranged.
Shortcut Method: Use the permutation formula. $$6P6 = \frac{6!}{(66)!} = \frac{6!}{0!} = 6! = 720$$ Therefore, there are 720 ways in which 6 letters of the word ‘MATHEMATICS’ can be arranged.
4. Find the number of ways in which 4 boys and 3 girls can be seated in a row if the girls are to be seated together.
Shortcut Method: Treat the 3 girls as one unit. So, we have 4 boys and 1 unit to arrange $$5P5 = \frac{5!}{(55)!} = \frac{5!}{0!} = 5! = 120$$ Therefore, there are 120 ways in which 4 boys and 3 girls can be seated in a row if the girls are to be seated together.
5. Find the number of 4digit numbers that can be formed using the digits 0, 1, 2, 3, 4, 5 if repetition of digits is not allowed.
Shortcut Method:

There are 6 choices for the first digit (excluding 0).

There are 5 choices for the second digit (excluding the digit chosen for the first digit).

There are 4 choices for the third digit (excluding the digits chosen for the first and second digits).

There are 3 choices for the fourth digit (excluding the digits chosen for the first, second, and third digits).
$$6 \times 5 \times 4 \times 3 = 360$$
Therefore, there are 360 ways to form a 4digit number without repeating any digits.
CBSE Class 11 and Class 12 Board Exams
1. Find the number of ways in which 3 letters can be chosen from the letters of the word ‘APPLE’.
Shortcut Method: Use the combination formula: $$nCr = \frac{n!}{r!(nr)!}$$ where n is the total number of items and r is the number of items to be selected. In this case, n = 5 and r = 3. $$5C3 = \frac{5!}{3!(53)!} = \frac{5!}{3!2!} = \frac{5 \times 4 \times 3!}{3!2!} = 10$$ Therefore, there are 10 ways in which 3 letters can be chosen from the letters of the word ‘APPLE’.
2. Find the number of ways in which 4 digits can be chosen from the digits 0, 1, 2, 3, 4, 5, 6, 7 if repetition of digits is allowed.
Shortcut Method:
 There are 8 choices for the first digit.
 There are 8 choices for the second digit.
 There are 8 choices for the third digit.
 There are 8 choices for the fourth digit.
$$8 \times 8 \times 8 \times 8 = 4096$$
Therefore, there are 4096 ways in which 4 digits can be chosen from the digits 0, 1, 2, 3, 4, 5, 6, 7 if repetition of digits is allowed.
3. Find the number of ways in which 5 students can be selected from a group of 10 students to form a committee if 2 particular students are not to be included.
Shortcut Method:
 There are 10 choices for the first student.
 There are 9 choices for the second student.
 There are 8 choices for the third student.
 There are 7 choices for the fourth student.
 There are 6 choices for the fifth student.
Subtract the number of ways the 2 particular students can be included in a group which is = $$2 C2 = 1$$
$$10 \times 9 \times 8 \times 7 \times 6 1 = 15119$$
Therefore, there are 15119 ways in which 5 students can be selected from a group of 10 students to form a committee if 2 particular students are not to be included.
4. Find the number of ways in which 6 books can be arranged on a shelf if 2 particular books are always to be kept together.
Shortcut Method: Treat the 2 particular books as a single unit. So, we have 5 books and 1 unit to be arranged. $$5P2 = \frac{5!}{(52)!} = \frac{5!}{3!} = 5 \times 4 \times 3! = 120$$ Therefore, there are 120 ways in which 6 books can be arranged on a shelf if 2 particular books are always to be kept together.
5. Find the number of 3digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if repetition of digits is not allowed.
Shortcut Method:
 There are 5 choices for the first digit.
 There are 4 choices for the second digit.
 There are 3 choices for the third digit.
$$5 \times 4 \times 3 = 60$$
Therefore, there are 60 ways in which 3digit numbers can be formed using the digits 1, 2, 3, 4, 5 if repetition of digits is not allowed.