mathematics-class-11-unit-08-chapter-05-permutation-combination-L-5-5-hhrtuaxxvho

### <Success>Solution</Success>
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<p>#$(B)=4 \times 3 \times 3 \times 2=72 .$</p>
<p>So the number of divisor of 12600 is 72 .</p>
<p><strong>(ii)</strong> $31752=2^3 \times 3^4 \times 7^2 $</p>
<p>$x=2^a \times 3^b \times 7^c, $</p>
<p>$B= \{(a, b, c): 0 \leq a \leq 3, \quad 0 \leq b \leq 4, \quad 0 \leq c \leq 2\}$</p>
<p>#$(B)=4 \times 5 \times 3=60 .$</p>
<p><strong>(iii)</strong>   $55125=3^2 \times 5^3 \times 7^2$</p>
<p>#$(B)=3 \times 4 \times 3=36$.</p>
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### <Success>Example 1</Success>
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<p>Eleven persons (including three specific persons $P, Q, R$ ) are to be seated on 11 seats so that $P, Q, R$ do not occupy adjacent seats. In how many ways this can be done?</p>
<p><strong>Solution:</strong>  First we arrange 8 persons in 8 ! ways. (excluding $P, Q, R$ )</p>
<p>_ (1) _ (2) _ (3) _ (4) _ (5) _ (6) _ (7) _ (8) _</p>
<p>In 9 remaining places, three seats can be placed in ${}^{9}P_{3}=9 \times 8 \times 7$ ways.</p>
<p>So $\text { total number of ways } =9 \times 8 \times 7 \times 8 ! $</p>
<p>$=20321280$</p>
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### <Success>Example 2</Success>
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<p>Six distinct symbols are transmitted through a communication channel. A total of 18 blanks are to be inserted between the symbols with at least two blanks between every pair of symbols. In how many ways symbols and blanks can be arranged?</p>
<p><strong>Solution:</strong>   Let the six symbols be</p>
<p>$S_1\cdots S_2\cdots S_3\cdots S_4\cdots S_5\cdots S_6$</p>
<p>So this can be considered as a problem of placing 18 identical balls (blanks) into 5 boxes. First we choose 2 blanks each in 1 way. Now remaining 8 blanks can be placed in 5 boxes in $\binom{8+5-1}{5-1} = \binom{12}{4}$ ways.</p>
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### <Success>Example 3</Success>
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<p>Find the number of non-ngative integer solutions to</p>
<p>$\quad quad x_1+x_2+x_3+x_4=10$.</p>
<p><strong>Solution:</strong>   This is same as distributing 10 identical balls in 4 distinct boxes :</p>
<p>$=\binom{10+4-1}{4-1}=\binom{13}{3} = 286$</p>
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<h3 id="successtheoremsuccess"><Success>Theorem</Success></h3>
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<p>Consider the equation,</p>
<p>$x_1+x_2+\cdots+x_n=r \ldots \ldots (1)$</p>
<p>where $r$ is a non-negative integer</p>
<p>(i) The number of non-negative integer solution to (1) is</p>
<p>$\binom{r+n-1}{r} = \binom{r+n-1}{n-1}$</p>
<p>(ii) The number of positive integer solutions to (i) is</p>
<p>$\binom{r-1}{n-1} = \binom{r-1}{r-n} \quad r \geq n$</p>
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### <Success>Example</Success>
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<p>Find the number of ways of distributing $r$ identical objects into $n$ distinct boxes such that Box 1 can hold at most one object.</p>
<p><strong>Solution:</strong>   If box 1 has no object, then we can distribute $r$ identical objects in $(n-1)$ boxes in $\binom{r+n-2}{r}$ ways. If box 1 has one object, then we can distribute $(r-1)$ identical objects in $(n-1)$ boxes in $\binom{r+n-3}{r-1}$ ways. So the total no. of ways</p>
<p>$\binom{r+n-2}{r} + \binom{r+n-3}{r-1}$.</p>
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### <Success>Proof</Success>
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<p>One can argue in the following way also:</p>
<p>One can place the first object into any $n$ boxes</p>
<p>One can place the second object into  $(n-1)$ boxes</p>
<p>One can place the $r^{th}$ object into $(n-r+1)$ boxes</p>
<p>So the total number of arrangements</p>
<p>$n \times(n-1) \cdots(n-r+1)=\frac{n !}{(n-r) !}={}^nP_{r} .$</p>
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### <Success>Further occupancy problems</Success>
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<p><strong>2.</strong>  The number of ways of distributing $r$ distinct objects into $n$ distinct boxes such that any box can hold any number of objects is $n^r$.</p>
<p><strong>Proof:</strong> Each of $r$ objects can be placed in $n$ possible boxes: $n \times n \times \cdots \times n=n^r$.</p>
<p><strong>3.</strong>  The number of ways of distributing $r$ distinct objects into $n$ distinct boxes such that the ordering of objects matters in each box is</p>
<p>$\binom{n+r-1}{r}r !=n(n+1) \cdots(n+r-1) .$</p>
<p><strong>Proof:</strong> Let us first assume that objects are identical. Then the number of ways of placing</p>
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