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The result of ^(n-1)Cr+^(n-1)C(r-1)=...

The result of `^(n-1)C_r+^(n-1)C_(r-1)=`

A

`^nC_(r-1)`

B

`^(n+1)C_r`

C

`^nC_r`

D

`^(n+1)C_(r-1)`

Text Solution

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The correct Answer is:
C
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PATHFINDER-PERMUTATION AND COMBINATIONS-QUESTION BANK
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