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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