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અહી $\left(\begin{array}{l}n \\ k\end{array}\right)$ એ ${ }^{n} C_{k}$ દર્શાવે છે અને $\left[\begin{array}{l} n \\ k \end{array}\right]=\left\{\begin{array}{cc}\left(\begin{array}{c} n \\ k \end{array}\right), & \text { if } 0 \leq k \leq n \\ 0, & \text { otherwise }\end{array}\right.$ છે.
જો $A_{k}=\sum_{i=0}^{9}\left(\begin{array}{l}9 \\ i\end{array}\right)\left[\begin{array}{c}12 \\ 12-k+i\end{array}\right]+\sum_{i=0}^{8}\left(\begin{array}{c}8 \\ i\end{array}\right)\left[\begin{array}{c}13 \\ 13-k+i\end{array}\right]$
અને $A_{4}-A_{3}=190 \mathrm{p}$ હોય તો $p$ ની કિમંત મેળવો.
$50$
$51$
$48$
$49$
Solution
$\mathrm{A}_{\mathrm{k}}=\sum_{\mathrm{i}=0}^{9}{ }^{9} \mathrm{C}_{\mathrm{i}}{ }^{12} \mathrm{C}_{\mathrm{k}-\mathrm{i}}+\sum_{\mathrm{i}=0}^{8}{ }^{8} \mathrm{C}_{\mathrm{i}}{ }^{13} \mathrm{C}_{\mathrm{k}-\mathrm{i}}$
$\mathrm{A}_{\mathrm{k}}={ }^{21} \mathrm{C}_{\mathrm{k}}+{ }^{21} \mathrm{C}_{\mathrm{k}}=2 \cdot{ }^{21} \mathrm{C}_{\mathrm{k}}$
$\mathrm{A}_{4}-\mathrm{A}_{3}=2\left({ }^{21} \mathrm{C}_{4}-{ }^{21} \mathrm{C}_{3}\right)=2(5985-1330)$
$190 \mathrm{p}=2(5985-1330) \Rightarrow \mathrm{p}=49$
