mathematics-class-11-unit-08-chapter-04-permutation-combination-L-4-5-mzersks9zvy

### <Success>Example 4</Success>
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<p>How many distinct terms are there in $x$ expressions</p>
<p>(i) $(x+y+z+t)(a+b+c+d+e)$</p>
<p>(ii) $\left(x_1+\cdots+x_m\right)\left(y_1+\cdots+y_n\right)\left(z_1+\cdots+z_t\right)$</p>
<p><strong>Solution:</strong><br>
(i) The total number of terms is exactly the same as in the cartesian product $A \times B$,</p>
<p>where $(A)=4,$ $ \quad (B)=5$</p>
<p>So $(A \times B)=4$ $\times 5=20$</p>
<p>(ii) Following the reasoning given in (i), we can say that the number of terms is m.n.t.</p>
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### <Success>Factorial notation</Success>
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<p>The notation $n$ ! represents He product of the first $n$ natural numbers</p>
<p>$n !=1 \times 2 \times 3 \times \cdots \times(n-1) \times n $</p>
<p>$ 1 !=1, \quad 2 !=1 \times 2=2, \quad  3 !=1 \times 2 \times 3=6$, and so on.</p>
<p>As a convention we define $0 !=1$.</p>
<p>$n !=n \times(n-1) \text { ! }$</p>
<p><strong>Remark:</strong> There are indications that ancient Indian mathematicians knew the concept product of consecutive natural numbers.‘a few hundred years B.C. The modern notations was introduced by Christian Cramp is 1808.<br>
We can notes that $n$ ! increases very rapidly as $n$ increases.
$
\begin{aligned}
& 4 !=24,\quad 5 !=120, \quad 6 !=720, \\
& 7 !=5040
\end{aligned}
$</p>
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### <Success>Theorem</Success>
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<p>The number of distinguishable permutations of $n$ objects of $k$ different types, where $n_1$ objects are alike ( $n_2$ are alike,… $n_k$) are alike and $n_1+n_2+\cdots+n_k=n$, is</p>
<p>$\frac{n !}{n_{1} ! n_{2} ! \cdots n_{k} !}$</p>
<p><strong>Example 1:</strong>  How many different 10-letter coded can be generated using 2 a’s, 3 b’s, 2 c’s and $3 d’s$ ?</p>
<p><strong>Solution:</strong>   The total number of distinct codes (10 letter)
$=\frac{10 !}{2 ! 3 ! 2 ! 3 !}=25200$</p>
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### <Success>Examples</Success>
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<p><strong>2.</strong> How many words can be made using all letters of the word ’ NATION’.</p>
<p><strong>Solution:</strong>   Here we have 6 letters where ’ $N$ ’ occurs twice. So the total number of arrangements of the six letters is $\frac{6 !}{2 !}=360$.</p>
<p><strong>3.</strong> How many distinct- 11 -letter arrangements
of the word ’ $P R O B A B I L I T Y$ ’ can be made?</p>
<p><strong>Solution:</strong>   Total number of letters is 11</p>
<p>$B \rightarrow$ occurs’ 2 times</p>
<p>$I \rightarrow$ occurs 2 times</p>
<p>So the total no. of arrangements is $\frac{11 !}{2 ! 2 !}=9979200$</p>
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### <Success>Example 4</Success>
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<p>In how many of these arrangements vowels appear together.</p>
<p><strong>Solution:</strong>   Here we have 7 consonants (B is repeated twice). Since all four vowels (I is repeaters) have to appear together, we have total 8 items. So the number of arrangements is</p>
<p>$\frac{8 !}{2 !} \text {. }$</p>
<p>Since vowels can be permuted among themselves, such arrangements are $\frac{4 !}{2 !}=12$
So the total number of words  from the letters of word ’ PROBABILITY’ where vowels occur together is</p>
<p>$\frac{8 !}{2 !} \times 12=241920 $</p>
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### <Success>Combinations</Success>
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<p>An unordered arrangement of $k$-distinct items from a set of $n$ distinct items is called a $k$-combination. The total number of all $k$-combinations is denoted by ${}^n C_k$ or $\left(\begin{array}{l}n_k\end{array}\right)$ ${}^n C_k$. $\quad(1 \leqslant k \leqslant n)$.</p>
<p>In order to evaluate $x={ }^n c_k$, we proceed as follows:</p>
<p>The number of ordered $k$-arrangements is ${}^n P_k$.</p>
<p>These are $k$ ! permutations of these $k$ objects. So</p>
<p>$\frac{{ }^n p_k}{k !}=x={ }^n c_k$</p>
<p>$\Rightarrow {}^n C_k=\frac{n !}{k !(n-k) !}, 1 \leqslant k \leqslant n$</p>
<p>Historically this egpe of formula appears in the work of Indiass Mathematicion Bhaskaracharya II ( 1114 - 1185 ).</p>
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