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UPSC NDA 2 2019 Mathematics Solved Paper

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Question : 30 of 120

Marks: +1, -0

What is

\binom{47}{4}+\binom{51}{3}+\binom{50}{3}

+\binom{51}{3}+\binom{50}{3}+\binom{49}{3}+\binom{48}{3}+\binom{47}{3}

C(47, 4)
C(52, 5)
C(52, 4)
C(47, 5)

Solution: 👈: Video Solution

Concept:

\binom{n}{r-1}+\binom{n}{r}=\binom{n+1}{r}

Calculation:

\Rightarrow \binom{47}{4}+\binom{51}{3}+\binom{50}{3}+\binom{49}{3}+\binom{48}{3}+\binom{47}{3}

=\left\{\binom{47}{4}+\binom{47}{3}\right\}+\binom{51}{3}+\binom{50}{3}+\binom{49}{3}+\binom{48}{3}

As we know that, C

(n, r-1)+\binom{n}{r}=\binom{n+1}{r}

\Rightarrow \left\{\binom{47}{4}+\binom{47}{3}\right\}+\binom{51}{3}+\binom{50}{3}+\binom{49}{3}+\binom{48}{3}

=\left\{\binom{48}{4}+\binom{48}{3}\right\}+\binom{51}{3}+\binom{50}{3}+\binom{49}{3}

\Rightarrow \left\{\binom{48}{4}+\binom{48}{3}\right\}+\binom{51}{3}+\binom{50}{3}+\binom{49}{3}

=\left\{\binom{49}{4}+\binom{49}{3}\right\}+\binom{51}{3}+\binom{50}{3}

\Rightarrow \left\{\binom{49}{4}+\binom{49}{3}\right\}+\binom{51}{3}+\binom{50}{3}

=\left\{\binom{50}{4}+\binom{50}{3}\right\}+\binom{51}{3}

\Rightarrow \left\{\binom{50}{4}+\binom{50}{3}\right\}+\binom{51}{3}

=\left\{\binom{51}{4}+\binom{51}{3}\right\}

=\binom{52}{4}

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