The number of ways of choosing 10 objects out of 31 objects of which 10 are identical and the remaining 21 are distinct, is

Options

A. `2^(20) - 1`
B. `2^(21)`
C. `2^(20)`
D. `2^(20) + 1`

Answer

C

Explanation

Given that, out of 31 objects 10 are identical and remaining 21 are distinct, so in following ways, we can choose 10 objects.
0 identical + 10 distincts, number of ways `= 1 xx ""^(21)C_(10)`
1 identical + 9 distincts, number of ways `= 1 xx ""^(21)C_(9)`
2 identicals + 8 distincts, number of ways `= 1 xx ""^(21)C_(8)`
........
........
So, total number of ways in which we can choose 10 objects is
`""^(21)C_(10) + ""^(21)C_(9) + ""^(21)C_(8) + ... + ""^(21)C_(0) = x ("let") " "...(i)`
`rArr ""^(21)C_(11) + ""^(21)C_(12) + ""^(21)C_(13) + ... + ""^(21)C_(21) = x " "...(ii)`
`[because ""^(n)C_(r) = ""^(n)C_(n-r)]`
On adding both Eqs. (i) and (ii) , we get
`2x = ""^(21)C_(0) + ""^(21)C_(1) + ""^(21)C_(2) + ... + ""^(21)C_(10) + ""^(21)C_(11) + ""^(21)C_(12) + ...+ ""^(21)C_(21)`
`rArr " " 2x = 2^(21) rArr " " x = 2^(20)`