mathematics-class-11-unit-06-chapter-06-probability-l-6-6-bt-ahkri76y

### <Success>Use of binomial theorem</Success>
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<p>But we have done it in a mechanical way:
such a solution $n$ not guaranteed to give all possible combination</p>
<p>The concept in “Generating Function” essentially a power series which may be finite or infinite.</p>
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### <Success>Example</Success>
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<p>In how many ways we can have two variables</p>
<p>$x \ \& \ y \ \Rightarrow \ 0 \leq x \leq 2 \quad$ & $\quad 1 \leq y \leq 2$</p>
<p>& $  {x+y}=3 \text { here } X \text { and } Y \text { are integer variables }$</p>
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<p>======</p>
<h3 id="successsolutionsuccess"><Success>Solution</Success></h3>
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<p>Consider the following two polynomial & compute their product.</p>
<p>$  \left(z^0+z^1+z^2\right)\left(z^1+z^2\right)$</p>
<p>$ = \left(1+z^2+ z^2\right)\left(z+z^2\right)$</p>
<p>$ = z+z^2+z^3+z^2+z^3+z^4$</p>
<p>$ = z+2 z^2+2 z^3+z^4 $</p>
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### <Success>Problem 4</Success>
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<p>$0<n_1<n_2<n_3<n_4<n_5, \quad \sum_{i=1}^5 n_i=20$</p>
<p>Note that all the 5 variables are integers.
Let</p>
<p>$ m_2=n_2-n_1$</p>
<p>$ m_3=n_3-n_2$</p>
<p>$ m_4=n_4-n_3$</p>
<p>$ m_5=n_5-n_4$</p>
<p>$\therefore n_1$ & $m_2, m_3, m_4, m_5$ are all $>0$</p>
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<p>======</p>
<h3 id="successsolutionsuccess-1"><Success>Solution</Success></h3>
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<p>More over</p>
<p>$n_1+n_2+n_3+n_n+n_5=20$</p>
<p>$ 5 n_1+4\left(n_2-n_1\right)+3\left(n_3-n_2\right)+2\left(n_4-n_3\right)+\left(n_5-n_4\right)=20$</p>
<p>or</p>
<p>$5 n_1+4 m_2+3 m_3+2 m_4+m_5=20$</p>
<p>when,</p>
<p>$n_1$ & all $m_i>0$</p>
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