====== Question 13 Exercise 6.2 ======
Solutions of Question 13 of Exercise 6.2 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.
=====Question 13(i)=====
Find the number of permutation of word "Excellence." How many of these permutations begin with $\mathrm{E}$ ?
====Solution====
The total number of letters in 'Excellence' are: $n=10$, out of which $m_1=4$ are $E, m_2=2$ are $L$ and $m_3=2$ are $C$.
Therefore,
\begin{align}\text{total number of permutations are}
&=\left(\begin{array}{c}
n \\
m_1, m_2, m_3
\end{array}\right)\\&=\left(\begin{array}{c}
10 \\
4,2,2
\end{array}\right) \\
& =\dfrac{10 !}{4 ! \cdot 2 ! \cdot 2 !}\\
&=37,800 \end{align}
Begin with $\mathrm{E}$
If we have to pick the combination of words that begin with $E$.
It means we have lixed the first one, and the remaining are
$n=9$ letters, out of which $m_1=3$ are $E$.
$m_2=2$ are $L$ and $m_3=2$ are $C$. Therefore,
\begin{align}\text{Number of permulations are}
&=\left(\begin{array}{c}
n \\
m_1, m_2, m_3
\end{array}\right)\\&=\dfrac{9 !}{3 ! \cdot 2 ! \cdot 2 !}\\
&=15,120 \end{align}
=====Question 13(ii)=====
Find the number of permutation of word "Excellence." How many of these permutations begin with $\mathrm{E}$ and end with $\mathrm{C}$ ?
====Solution====
The total number of letters in 'Excellence' are: $n=10$,
out of which $m_1=4$ are $E$,
$m_2=2$ are $L$
$m_3=2$ are $C$.
Therefore,
\begin{align}\text{total number of permutations are}
& =\left(\begin{array}{c}
n \\
m_1, m_2, m_3
\end{array}\right)\\&=\left(\begin{array}{c}
10 \\
4,2,2
\end{array}\right) \\
& =\dfrac{10 !}{4 ! \cdot 2 ! \cdot 2 !}\\
&=37,800\end{align}
Begin with $E$ and end with $C$
If begin with $E$ and end with $C$, means we fixed the two letters and two place.
So, the remaining letters are $n=8$,
out of which $m_1=3$ are $E$, $m_2=2$ are $L$ and $m_3=1$ are $C$.
$\therefore$
\begin{align}\text{Number of permutations are}& =\left(\begin{array}{c}
n \\
m_1, m_2,m_3
\end{array}\right)\\&=\left(\begin{array}{c}
8 \\
3.2 .1
\end{array}\right) \\
& =\dfrac{8 !}{3 ! \cdot 2 ! .1 !}\\
&=3360 \end{align}
=====Question 13(iii)=====
Find the number of permutation of word "Excellence." How many of these permutations begin with $\mathrm{E}$ and end with $\mathrm{E}$ ?
====Solution====
The total number of letters in 'Excellence' are: $n=10$,
out of which $m_1=4$ are $E, m_2=2$ are $L$ and $m_3=2$ are $C$.
Therefore,
\begin{align}\text{total number of permutations are}& =\left(\begin{array}{c}
n \\
m_1, m_2, m_3
\end{array}\right)\\&=\left(\begin{array}{c}
10 \\
4,2,2
\end{array}\right) \\
& =\dfrac{10 !}{4 ! \cdot 2 ! \cdot 2 !}\\
&=37,800 \end{align}
Begin with $E$ and end with $E$
If first and last place is filled with $E$,
the remaining letters are $n=8$,
out of which $m_1=1$ are $E$
$m_2=2$ are $L$ and $m_3=2$ are $C$.
\begin{align}\text{Numbers of perinutations are}& =\left(\begin{array}{c}
n \\
m_1, m_2, m_3
\end{array}\right)\\&=\left(\begin{array}{c}
8 \\
1,2,1
\end{array}\right) \\
& =\dfrac{8 !}{1 ! \cdot 2 ! \cdot 2 !}\\
&=10,080\end{align}
=====Question 13(iv)=====
Find the number of permutation of word "Excellence." How many of these permutations do not begin with $\mathrm{E}$ ?
====Solution====
The total number of letters in 'Excellence' are: $n=10$,
out of which $m_1=4$ are $E, m_2=2$ are $L$ and $m_3=2$ are $C$.
Therefore,
\begin{align}\text{total number of permutations are}& =\left(\begin{array}{c}
n \\
m_1, m_2, m_3
\end{array}\right)\\&=\left(\begin{array}{c}
10 \\
4,2,2
\end{array}\right) \\
& =\dfrac{10 !}{4 ! \cdot 2 ! \cdot 2 !}\\
&=37,800\end{align}
Do not begin with $\mathrm{E}$
\begin{align}\text{Number of permutations}& = \text{Total permutations} - {Numbe\,of\, permutation\, begin\, with\,} \mathrm{E}\\
&=37,800-15,120\\
&=22,680 \end{align}
=====Question 13(v)=====
Find the number of permutation of word "Excellence." How many of these permutations contain two 2L's together?
====Solution====
The total number of letters in 'Excellence' are:$n=10$,
out of which $m_1=4$ are $E, m_2=2$ are $L$ and $m_3=2$ are $C$.
Therefore, \begin{align}\text{total number of permutations are}
& =\left(\begin{array}{c}
n \\
m_1, m_2, m_3
\end{array}\right)\\&=\left(\begin{array}{c}
10 \\
4,2,2
\end{array}\right) \\
& =\dfrac{10 !}{4 ! \cdot 2 ! \cdot 2 !}\\
&=37,800 \end{align}
Contain two $2$L's together
If two $2 L^{\prime} s$ are to be kept together,
then we shall deal these two letters as single,
the remaining are $n=9$ out of which $m_1=4$ are $E$,
$m_2=$ 2 are $C$. Therefore,
\begin{align}\text{Number of permutations are}
& =\left(\begin{array}{c}
n \\
m_1, n_2
\end{array}\right)\\&=\left(\begin{array}{c}
9 \\
4,2
\end{array}\right) \\
& =\dfrac{9 !}{4 ! .2 !}\\
&=7,560 \end{align}
=====Question 13(vi)=====
Find the number of permutation of word "Excellence." How many of these permutations do not contain 2L's together?
====Solution====
The total number of letters in 'Excellence' are: $n=10$,
out of which $m_1=4$ are $E, m_2=2$ are $L$ and $m_3=2$ are $C$.
Therefore, \begin{align}\text{total number of permutations are}
& =\left(\begin{array}{c}
n \\
m_1, m_2, m_3
\end{array}\right)\\&=\left(\begin{array}{c}
10 \\
4,2,2
\end{array}\right) \\
& =\dfrac{10 !}{4 ! \cdot 2 ! \cdot 2 !}\\
&=37,800 \end{align}
Do not contain $2 L^{\prime} s$ together
In this case,
\begin{align}\text{the total number of permutations are}&=\text{total permutations} -\quad \text{Permutations containing}\, 2 L's together\\
&=37800-7,560\\
&=30.240\end{align}
====Go To====
[[math-11-kpk:sol:unit06:ex6-2-p8 |< Question 12 ]]
[[math-11-kpk:sol:unit06:ex6-2-p10|Question 14 & 15 >]]