**Example 1:** **(a)** How many words can be formed using the letters of the word TRIANGLES? **(b)** How many of these words start with T and end with S?

**Solution: (a) **There are 9 distinct letters in the given word. Thus, the number of different permutations (or arrangements) of the letters of this word is \(^9{P_9} = 9!\).

**(b)** If we fix T at the start and S at the end of the word, we have to *permute* 7 distinct letters in 7 places. This can be done in \(^7{P_7} = 7!\) ways. Thus, the number of such words is 7!.

**Example 2: **How many different 5-letter words can be formed using the letters from A to J (total ten letters) such that each word has at least one letter repeated?

**Solution:** We count:

(i) the total number of 5-letter words (with or without repetition) which can be formed using 10 distinct letters. This number is \({10^5}\), since each place in the 5-letter word can be filled in 10 different ways, and so, the required number is \(10 \times 10 \times 10 \times 10 \times 10 = {10^5}\).

(ii) the number of 5-letter words which can be formed using 10 distinct digits, such that no digit is repeated. This number is \(^{10}{P_5}\).

The answer to the original question is the difference of these two numbers. If you take out, from the set of all the words, those words which have no repetition of letters, then you get the set of words which have at least one letter repeated. Thus, the final answer is \({10^5}{ - ^{10}}{P_5}\).

**Example 3:** There are 10 students in a class. Two of these students are Nadeesh and Kopal, who don’t get along very well. In how many ways can the teacher arrange the students in a row, so that Nadeesh and Kopal are not together?

**Solution: **The total number of arrangements of the 10 students is \(^{10}{P_{10}} = 10!\).

Now, let us count the number of arrangements in which Nadeesh and Kopal are together. Treating Nadeesh + Kopal as a single unit (call it NK), we have a total of 9 entities which we can permute: the 8 students, and NK. These 9 entities can be permuted in \(^9{P_9} = 9!\) ways. But for each of 9! permutations, N and K can be permuted among themselves in 2! or 2 ways: NK or KN. Thus, the total number of permutations in which Nadeesh and Kopal are together, is \(2 \times 9!\).

We conclude that the number of permutations in which Nadeesh and Kopal are not together, is

\[10! - \left( {2 \times 9!} \right)\]

**Example 4:** There is a row of *r* chairs. A group of *n* people needs to be seated in this row, where *n* is less than *r*. How many different permutations are possible?

**Solution: ** Earlier, when we talked about seating *n* people in a row of *r* chairs, we calculated the number of permutations of the *n* people taken *r* at a time as \(^n{P_r}\).

However, the situation here is different. The number of chairs is more than the number of people. Also, in counting a permutation as distinct from others, not only does the order of seating matter, but what also matters is – which person sits in which chair?

For example, suppose that there are 4 people (A, B, C and D), and 6 chairs. Consider the following two arrangements:

\[\begin{align}&{\fbox{B}}\;\;{\fbox{ }}\;\;{\fbox{A}}\;\;{\fbox{C}}\;\;{\fbox{ }}\;\;{\fbox{D}}\\& {\fbox{ }}\;\;{\fbox{B}}\;\;{\fbox{A}}\;\;{\fbox{ }}\;\;{\fbox{C}}\;\;{\fbox{D}} \end{align}\]

Even though the ordering of the people is the same (B, A, C, D), the chairs occupied in the two cases are different, and hence these will be counted as different arrangements.

The required number of permutations can be obtained as follows: rather than arranging *people onto chairs*, think of arranging *chairs onto people*. Thus, you have to calculate the permutations of *r* chairs, taken *n* at a time (because there are *n* people). The answer will be \(^r{P_n}\).

Before proceeding further, make sure you understand the logic behind this answer.

**Example 5:** 7 boys and 5 girls need to be seated in a row, so that the no two girls sit together. In how many ways can this be done?

**Solution:**Let us first arrange the 7 boys in a row. This can be done in \(^7{P_7} = 7!\) ways. Now, observe that 8 *spaces* are created by these 7 boys. In the following figure, B represents a boy, while a box represents a *space*:

\[\boxed{\,\,\,\,\,}\,\,B\,\,\boxed{\,\,\,\,\,}\,\,B\,\,\boxed{\,\,\,\,\,}\,\,B\,\,\boxed{\,\,\,\,\,}\,\,B\,\,\boxed{\,\,\,\,\,}\,\,B\,\,\boxed{\,\,\,\,\,}\,\,B\,\,\boxed{\,\,\,\,\,}\,\,B\,\,\boxed{\,\,\,\,\,}\]

Each space can be occupied by a girl. If every girl occupies one of these spaces, then we can be assured that no two girls will be together. The 5 girls can be arranged in the 8 spaces in \(^8{P_5}\) ways.

The final answer will be the product of these two numbers, the number of ways of arranging the 7 boys, and then the number of ways of arranging the 5 girls in the 8 spaces thus created: \(7!\,\,{ \times ^8}{P_5}\).