Question

a) How many different strings can be made from the word PEPPERCORN when all the letters are used?

b) How many of these strings start and end with the letter P?

c) In how many of these strings are the three letter Ps consecutive?

Short answer

(a) There are \({\rm{151,200}}\)different strings can be made from the word PEPPERCORN when all the letters are used.

(b) There are \({\rm{10,080}}\) different strings start and end with the letter P.

(c) In \({\rm{512 10,080}}\) different strings are the three letters Ps consecutive.

Step-by-step solution

Introduction

The product rule (also known as the Leibniz rule or Leibniz product rule) is a formula for calculating the derivatives of two or more functions in calculus.

Explanation

a)

Let us simplify,

\(\begin{array}{l}{\rm{n = 10}}\\{\rm{k = 6}}\\{{\rm{n}}_{\rm{1}}}{\rm{ = 3}}\\{{\rm{n}}_{\rm{2}}}{\rm{ = 2}}\\{{\rm{n}}_{\rm{3}}}{\rm{ = 2}}\\{{\rm{n}}_{\rm{4}}}{\rm{ = 1}}\\{{\rm{n}}_{\rm{5}}}{\rm{ = 1}}\\{{\rm{n}}_{\rm{6}}}{\rm{ = 1}}\end{array}\)

Distributing\({\rm{n}}\)distinguishable objects into\({\rm{k}}\)distinguishable boxes such that\({{\rm{n}}_{\rm{i}}}\)objects are placed in box\({\rm{i(i = 1,2,3,4,5)}}\)can be done in\(\frac{{{\rm{n!}}}}{{{{\rm{n}}_{\rm{1}}}{\rm{!}}{{\rm{n}}_{\rm{2}}}{\rm{! \ldots }}{{\rm{n}}_{\rm{k}}}{\rm{!}}}}\)ways,

\(\frac{{{\rm{10!}}}}{{{\rm{3!2!2!1!1!1!}}}}{\rm{ = 151,200}}\).

Hence, there are \({\rm{151,200}}\)different strings can be made from the word PEPPERCORN when all the letters are used.

Explanation

(b)

If the first and last letter of the string is a \({\rm{P}}\), then there are \({\rm{8}}\) letters remaining of which \({\rm{1}}\) is a \({\rm{P,2E}}\) 's, \({\rm{2}}\)R's, \({\rm{1C,1O}}\) and\({\rm{1\;N}}\).

\({\rm{n = 8}}\)

\(\begin{array}{l}{\rm{k = 6}}\\{{\rm{n}}_{\rm{1}}}{\rm{ = 1}}\\{{\rm{n}}_{\rm{2}}}{\rm{ = 2}}\\{{\rm{n}}_{\rm{3}}}{\rm{ = 2}}\\{{\rm{n}}_{\rm{4}}}{\rm{ = 1}}\end{array}\)

\(\begin{array}{l}{{\rm{n}}_{\rm{5}}}{\rm{ = 1}}\\{{\rm{n}}_{\rm{6}}}{\rm{ = 1}}\end{array}\)

Distributing \({\rm{n}}\) distinguishable objects into \({\rm{k}}\) distinguishable boxes such that \({{\rm{n}}_{\rm{i}}}\) objects are placed in box \({\rm{i(i = 1,2,3,4,5)}}\) can be done in:

\(\frac{{{\rm{n!}}}}{{{{\rm{n}}_{\rm{1}}}{\rm{!}}{{\rm{n}}_{\rm{2}}}{\rm{!}}...{{\rm{n}}_{\rm{k}}}{\rm{!}}}}\)Ways,

\(\frac{{{\rm{8!}}}}{{{\rm{1!2!2!1!1!1!}}}}{\rm{ = 10,080}}\).

Therefore, there are \({\rm{10,080}}\) different strings start and end with the letter P.

Explanation

(c)

The three consecutive P's can occur in 8 ways (the dots. represent an unknown letter).

\(\begin{array}{l}PPP \ldots \ldots .\\.PPP \ldots \ldots \\..PPP \ldots .\\ \ldots PPP \ldots \\ \ldots PPP \ldots \\ \ldots ..PPP..\\ \ldots ...PPP.\\ \ldots ....PPP\end{array}\)

There are \({\rm{7}}\) letters remaining of which \({\rm{2}}{{\rm{E}}^{\rm{\cent}}}{\rm{s,2}}{{\rm{R}}^{\rm{\cent}}}{\rm{s,1C,1O and 1\;N}}{\rm{.}}\)

\({\rm{n - 7}}\)

\({\rm{k - 5}}\)

\({{\rm{n}}_{\rm{2}}}{\rm{ - 2}}\)

\(\begin{array}{l}{{\rm{n}}_{\rm{3}}}{\rm{ - 2}}\\{{\rm{n}}_{\rm{4}}}{\rm{ - 1}}\end{array}\)

\(\begin{array}{l}{{\rm{n}}_{\rm{5}}}{\rm{ - 1}}\\{{\rm{n}}_{\rm{6}}}{\rm{ - 1}}\end{array}\)

Distributing \({\rm{n}}\) distinguishable objects into \({\rm{k}}\) distinguishable boxes such that \({{\rm{n}}_{\rm{i}}}\) objects are placed in box \({\rm{i(i - 1,2,3,4,5)}}\) can be done in:

\(\frac{{{\rm{n!}}}}{{{{\rm{n}}_{\rm{1}}}{\rm{!}}{{\rm{n}}_{\rm{2}}}{\rm{!}}...{{\rm{n}}_{\rm{k}}}{\rm{!}}}}\) ways.

\(\frac{{{\rm{7!}}}}{{{\rm{2!2!1!1!1!}}}}{\rm{ - 1260}}\)

Use the product rule:

\({\rm{8 \times 1260 - 10,080}}\).

Thus, in \({\rm{512 10,080}}\) different strings are the three letters Ps consecutive.