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Prove that combinatorial argument that `""^(n+1)C_r=^n C_r+^n C_(r-1)""`

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Consider n+1 different toys Then,
(no. of ways of selecting r toys out of (n+1) different toys)
(no. of ways of selecting r toys out of (n+1) different toys)
=(no. of ways of selecting r toys when `T_(0)` is always included)
+ (no. of ways of selecting r toys when `T_(0)` is always excluded)
`=""^(n)C_(r-1)+ ""^(n)C_(r )`
Hence, `""^(n+1)C_(r )= ""^(n)C_(r-1)+ ""^(n)C_(r )`.
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