Permutations and Combinations 2 Question 1
1. The number of ways of choosing 10 objects out of 31 objects of which 10 are identical and the remaining 21 are distinct, is
(2019 Main, 12 April I)
(a) $2^{20}-1$
(b) $2^{21}$
(c) $2^{20}$
(d) $2^{20}+1$
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Answer:
Correct Answer: 1. (c)
Solution:
- 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 $=1 \times{ }^{21} C _{10}$
1 identical +9 distincts, number of ways $=1 \times{ }^{21} C _9$
2 identicals +8 distincts, number of ways $=1 \times{ }^{21} C _8$
. . . . . .
So, total number of ways in which we can choose 10 objects is
${ }^{21} C _{10}+{ }^{21} C _9+{ }^{21} C _8+\ldots+{ }^{21} C _0=x$ (let)
$\Rightarrow{ }^{21} C _{11}+{ }^{21} C _{12}+{ }^{21} C _{13}+\ldots+{ }^{21} C _{21}=x$
On adding both Eqs. (i) and (ii), we get
$$ \left[\because{ }^{n} C _r={ }^{n} C _{n-r}\right] $$
$$ \begin{aligned} & 2 x={ }^{21} C _0+{ }^{21} C _1+{ }^{21} C _2 \\ &+\ldots+{ }^{21} C _{10} \\ &+{ }^{21} C _{11}+{ }^{21} C _{12}+\ldots+{ }^{21} C _{21} \\ & \Rightarrow \quad 2 x=2^{21} \Rightarrow \quad x=2^{20} \end{aligned} $$
