${ }^{37} C_4+\sum_{r=1}^5(42-r)_{C_r}=$
${ }^{37} C_4+\sum_{r=1}^5(42-r)_{C_r}=$
- ${ }^{41} C_4$
- ${ }^{39} C_4$
- ${ }^{38} C_4$
- ${ }^{42} C_4$
Solution
Given that,
$
\begin{aligned}
& { }^{37} C_4+\sum_{r=1}^5{ }^{(42-r)} C_r \\
& ={ }^{37} C_4+{ }^{41} C_1+{ }^{40} C_2+{ }^{39} C_3+{ }^{38} C_3+{ }^{37} C_3 \\
& ={ }^{37} C_3+{ }^{37} C_4+{ }^{38} C_3+{ }^{39} C_3+{ }^{40} C_3+{ }^{41} C_3
\end{aligned}
$
As we know,
$
\begin{aligned}
& { }^n C_{r-1}+{ }^n C_r={ }^{n+1} C_r \\
& ={ }^{38} C_4+{ }^{38} C_3+{ }^{39} C_3+{ }^{40} C_3+{ }^{41} C_3 \\
& ={ }^{39} C_4+{ }^{39} C_3+{ }^{40} C_3+{ }^{41} C_3 \\
& ={ }^{40} C_4+{ }^{40} C_3+{ }^{41} C_3 \\
& ={ }^{41} C_4+{ }^{41} C_3={ }^{42} C_4
\end{aligned}
$
Asked in: AP EAMCET 2017 (26 Apr Shift 1)
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