Question 12 Exercise 6.2

Solutions of Question 12 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.

How many different word can be formed from the letters “BOOKWORM” if the letters are taken all at a time?

BOOKWORM
The total number of letters in word BOOKWORM are $8.$

$n=8$ out of which three are $\mathrm{O}$,

so $m_1=3$..

Thus total number of different words using all at a time are: \begin{align} \left(\begin{array}{c} n \\ m 1 \end{array}\right)&=\left(\begin{array}{l} 8 \\ 3 \end{array}\right) \\ & =\dfrac{8 !}{3 !}\\ &=\dfrac{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 !}{3 !}\\ &=6,720 \end{align}

How many different word can be formed from the letters “BOOKKEEPER” if the letters are taken all at a time?

BOOKKEEPER
The total number of letters in $\mathrm{BOOK}$ KEEPER are ten.

$n=10$, out of which two are $\mathrm{O}$,

so $m_1=2$, three are $\mathrm{E}$,

so $m_2=3$, two are $\mathrm{K}$,

so $m_3=2$. Thus the total number of different words are: \begin{align} \left(\begin{array}{c} n \\ m_1, m_2, m_3 \end{array}\right)&=\left(\begin{array}{c} 10 \\ 2,3.2 \end{array}\right) \\ & =\dfrac{10 !}{2 ! \cdot 3 ! \cdot 2 !}\\ &=151,200 \end{align}

How many different word can be formed from the letters “ABBOTTABAD” if the letters are taken all at a time?

ABBOTABAD
Total number of letters are ten, so $n=10$ out of which three are $\mathrm{A}$,

so $m_1=3$, three are $B$, so $m_2=3$, and two are $T$,

so $m_3=2$.

Thus the total number of different words formed are: \begin{align} \left(\begin{array}{c} n \\ m_1, m_2, m_3 \end{array}\right)&=\left(\begin{array}{c} 10 \\ 3,3,2 \end{array}\right) \\ & =\dfrac{10 !}{3 ! \cdot 3 ! \cdot 2 !}\\ &=50,400 \end{align}

How many different word can be formed from the letters “LETTER” if the letters are taken all at a time?

LETTER
The total number of letters in letter are six.

so, $n=6$ out of which two are t,

so $m_1=2$ and two are e, so $m_2=2$.

Thus the total number of different words formed are: \begin{align}\left(\begin{array}{c} n \\ m_1, m_2 \end{array}\right)&=\left(\begin{array}{c} 6 \\ 2,2 \end{array}\right)\\&=\dfrac{6 !}{2 ! \cdot 2 !}\\ &=180 \end{align}

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