Question

The number of ways of choosing 10 objects out of 31 objects of which 10 are identical and the remaining 21 are distinct, is?
A. ${{2}^{20}}$
B. ${{2}^{20}}-1$
C. ${{2}^{20}}+1$
D. ${{2}^{21}}$

Answer 1

Hint: In permutations the order does matter but whereas in combinations the order doesn’t matter. A combination is a mathematical technique that determines the number of possible arrangements in a collection of items where the order of the selection does not matter.
The number of ways to select any number of objects from n or more identical objects is 1.
The number of ways to select n objects from r distinct objects without repetition would be ${}^{n}{{C}_{r}}=\dfrac{n!}{(n-r)!r!}$ .

Complete step by step answer:
By observing the question we have that there are total of 31 objects with 10 identical and 21 distinct and we should select 10 objects
Number of ways for selecting n objects from 21 identical objects = ${}^{21}{{C}_{n}}$
Number of ways for selecting any number of objects from 10 identical objects = 1.
The possible combinations would be
0 identical 10 distinct = $1\times ({}^{21}{{C}_{10}})={}^{21}{{C}_{10}}$
1 identical 9 distinct = $1\times ({}^{21}{{C}_{9}})={}^{21}{{C}_{9}}$
2 identical 8 distinct = $1\times ({}^{21}{{C}_{8}})={}^{21}{{C}_{8}}$
3 identical 7 distinct = $1\times ({}^{21}{{C}_{7}})={}^{21}{{C}_{7}}$
.
.
.
.
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9 identical, 1 distinct = $1\times ({}^{21}{{C}_{1}})={}^{21}{{C}_{1}}$
10 identical 0 distinct = $1\times ({}^{21}{{C}_{0}})={}^{21}{{C}_{0}}$
The total number of combinations = $\left( {}^{21}{{C}_{10}}+{}^{21}{{C}_{9}}+{}^{21}{{C}_{8}}+.............+{}^{21}{{C}_{1}}+{}^{21}{{C}_{0}} \right)$
We also know that, ${}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}$
Let us assume that $\left( {}^{21}{{C}_{11}}+{}^{21}{{C}_{12}}+{}^{21}{{C}_{13}}+.............+{}^{21}{{C}_{20}}+{}^{21}{{C}_{21}} \right)=x$---eq(i)
By using ${}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}$ which implies that ${}^{21}{{C}_{10}}={}^{21}{{C}_{11}},{}^{21}{{C}_{9}}={}^{21}{{C}_{12}},{}^{21}{{C}_{8}}={}^{21}{{C}_{13}},................,{}^{21}{{C}_{1}}={}^{21}{{C}_{20}},{}^{21}{{C}_{0}}={}^{21}{{C}_{21}}$
We can say that $\left( {}^{21}{{C}_{10}}+{}^{21}{{C}_{9}}+{}^{21}{{C}_{8}}+.............+{}^{21}{{C}_{1}}+{}^{21}{{C}_{0}} \right)=x$-----eq(ii)
Adding eq(i) and eq(ii)
$2x=\left( {}^{21}{{C}_{0}}+{}^{21}{{C}_{1}}+{}^{21}{{C}_{2}}+{}^{21}{{C}_{3}}.............+{}^{21}{{C}_{10}}+{}^{21}{{C}_{11}}+{}^{21}{{C}_{12}}+{}^{21}{{C}_{13}}+......{}^{21}{{C}_{21}} \right)$
From binomial expansion we know that,
Replacing x=1 in the expansion\[{{\left( 1+x \right)}^{n}}~={{~}^{n}}{{C}_{0}}~+{{~}^{n}}{{C}_{1}}~x\text{ }+{{~}^{n}}{{C}_{2}}~{{x}^{2}}~+...+{{~}^{n}}{{C}_{x}}~{{x}^{n}}\] .
\[\Rightarrow {{2}^{n}}~={{~}^{n}}{{C}_{0}}~+{{~}^{n}}{{C}_{1}}~x\text{ }+{{~}^{n}}{{C}_{2}}~+...+{{~}^{n}}{{C}_{n}}.\]
$\Rightarrow \left( {}^{21}{{C}_{0}}+{}^{21}{{C}_{1}}+{}^{21}{{C}_{2}}+{}^{21}{{C}_{3}}.............+{}^{21}{{C}_{10}}+{}^{21}{{C}_{11}}+{}^{21}{{C}_{12}}+{}^{21}{{C}_{13}}+......{}^{21}{{C}_{21}} \right)$=$\left( {}^{21}{{C}_{0}}+{}^{21}{{C}_{1}}+{}^{21}{{C}_{2}}+..............{}^{21}{{C}_{21}} \right)$
By using the above expansion,
$\left( {}^{21}{{C}_{0}}+{}^{21}{{C}_{1}}+{}^{21}{{C}_{2}}+..............{}^{21}{{C}_{21}} \right)={{2}^{21}}$
$\begin{align}
  & 2x={{2}^{21}} \\
 & x={{2}^{20}} \\
\end{align}$
As we have considered the total number of combinations to be x
Therefore, $x={{2}^{20}}$

So, the correct answer is “Option A”.

Note: Students should read the question properly as it mentioned that there are identical objects and not all are distinct objects. Some of the standard binomial expansions should be remembered to solve problems of this kind easily. Sometimes it might be asked to even arrange the combinations selected for that we should use permutations.