# 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}

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