AP EAMCET · Maths · Binomial Theorem
The value of \(\Sigma_{r=0}^4{ }^{(19-r)} C_3+{ }^{15} C_4\) is equal to
- A \({ }^{21} C_4\)
- B \({ }^{19} \mathrm{C}_4\)
- C \({ }^{20} C_4\)
- D \({ }^{16} \mathrm{C}_4\)
Answer & Solution
Correct Answer
(C) \({ }^{20} C_4\)
Step-by-step Solution
Detailed explanation
\(\sum_{r=0}^4{ }^{19-r} C_3+{ }^{15} C_4\) \(={ }^{19} C_3+{ }^{18} C_3+{ }^{17} C_3+{ }^{16} C_3+{ }^{15} C_3+{ }^{15} C_4\)…
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