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7.Binomial Theorem
normal
જો $\sum\limits_{K = 1}^{12} {12K{.^{12}}{C_K}{.^{11}}{C_{K - 1}}} $ ની કિમત $\frac{{12 \times 21 \times 19 \times 17 \times ........ \times 3}}{{11!}} \times {2^{12}} \times p$ હોય તો $p$ ની કિમત મેળવો
A
$2$
B
$4$
C
$8$
D
$6$
Solution
$ = 12k \cdot \frac{{{{12}^{11}}}}{k}{C_{k – 1}}{\,^{11}}{C_{k – 1}}$
$\sum\limits_{K = 1}^n {12.K} {\,^{12}}{C_K}.{\,^{11}}{C_{K – 1}} = {12^2}\sum\limits_{K = 1}^{12} {{{\left( {^{11}{C_{K – 1}}} \right)}^2}} $
$=12^{2} \cdot \frac{22 !}{11 ! 11 !}$
$=12 \cdot \frac{21.19 .17 \ldots .3}{11 !} 2^{12} .6 \Rightarrow p=6$
Standard 11
Mathematics
