Question

a) Derive a formula for the number of permutations ofobjects of k different types, where there are${\mathbf{n}}_{\mathbf{1}}$ indistinguishable objects of type one,${\mathbf{n}}_{\mathbf{2}}$ indistinguishable objects of type two,..., and${\mathbf{n}}_{\mathit{k}}$ indistinguishable objects of type k.

b) How many ways are there to order the letters of the word INDISCREETNESS?

Short answer

(a) A formula for the number of permutations of objects are$\frac{\mathrm{n}!}{{\mathrm{n}}_1!{\mathrm{n}}_2!\dots {\mathrm{n}}_{\mathrm{k}}!}$ways.

(b) The number of ways that are there to order the letters of the word INDISCREETNESS is $605,404,800$ways .

Step-by-step solution

Concept Introduction

Permutation is a method of arranging objects in a specific order. When working with permutation, it's important to think about both selection and layout. In a nutshell, ordering is critical in permutations. To put it another way, a permutation is an ordered combination.

Deriving a Formula

(a)

Distributing distinguishable objects into distinguishable boxes such that ${\mathrm{n}}_{\mathrm{i}}$objects are place in box $\left(\mathrm{i}=1,2,3,4,5\right)$can be done in role="math" localid="1668689790149" $\frac{\mathrm{n}!}{{\mathrm{n}}_1!,{\mathrm{n}}_2!...{\mathrm{n}}_{\mathrm{k}}!}$ways.

Therefore, the result is obtained as$\frac{\mathrm{n}!}{{\mathrm{n}}_1!,{\mathrm{n}}_2!...{\mathrm{n}}_{\mathrm{k}}!}$ .

Finding the number of ways

(b)

INDISCREETNESS contains in total letters of which 2 l's, 2 N's, 1 D, 3 S's, 1 C, 1 R, 3 E's and 1 T.

Here, it can be seen that –$-\mathrm{n}=14,\mathrm{k}=8,{\mathrm{n}}_1=2,{\mathrm{n}}_2=2,{\mathrm{n}}_3=1,{\mathrm{n}}_4=3,{\mathrm{n}}_5=1,{\mathrm{n}}_8=1,{\mathrm{n}}_7=3$ and${\mathrm{n}}_8=1$ .

Distributing n distinguishable objects intodistinguishable boxes such that${\mathrm{n}}_{\mathrm{i}}$objects are place in box i$\left(\mathrm{i}=1,2,3,4,5\right)$ can be done in$\frac{\mathrm{n}!}{{\mathrm{n}}_1!,{\mathrm{n}}_2!...{\mathrm{n}}_{\mathrm{k}}!}$ways.

Substituting the values, it is obtained –

$$\begin{gathered} \frac{14!}{2!2!1!3!1!1!3!1!}=\frac{14!}{(1!{)}^4(2!{)}^2(3!{)}^2} \\ =605,404,800 \end{gathered}$$

Therefore, the result is obtained as 605,404,800 .