What is {}^{47}C₄ + {}^{51}C₃ + _{{j}=2}^{5} {}^{52-j}C₃ equal to?

Correct Answer is : 52C4{}^{52}C₄52C4

Correct Answer is : 52C4{}^{52}C₄52C4

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UPSC NDA 2 2016 Mathematics Solved Paper

Question : 15 of 120

Marks: +1, -0

What is

{}^{47}C_{4} + {}^{51}C_{3} + \sum\limits_{\mathrm{j}=2}^{5} {}^{52-j}C_{3}

equal to?

Solution:

Concept:

The addition rule of combinations:

{}^{n}C_{r} + {}^{n}C_{r+1} = {}^{n+1}C_{r+1}

Explanation:

First, expand the summation:

\sum_{j=2}^{5} {}^{52-j}C_{3} = {}^{50}C_{3} + {}^{49}C_{3} + {}^{48}C_{3} + {}^{47}C_{3}

Now rewrite the full expression:

{}^{47}C_{4} + {}^{51}C_{3} + ({}^{50}C_{3} + {}^{49}C_{3} + {}^{48}C_{3} + {}^{47}C_{3})

Group

{}^{47}C_{4}

and

{}^{47}C_{3}

({}^{47}C_{3}+{}^{47}C_{4}) = {}^{48}C_{4}

Add

{}^{48}C_{3}

{}^{48}C_{4}+{}^{48}C_{3}={}^{49}C_{4}

Add

{}^{49}C_{3}

{}^{49}C_{4}+{}^{49}C_{3}={}^{50}C_{4}

Add

{}^{50}C_{3}

{}^{50}C_{4}+{}^{50}C_{3}={}^{51}C_{4}

Finally, add

{}^{51}C_{3}

{}^{51}C_{4}+{}^{51}C_{3}={}^{52}C_{4}

Thus the entire sum simplifies to

{}^{52}C_{4}

Answer:

{}^{52}C_{4}

(Option A).

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