Understanding the Problem Setup

The question asks for the number of different ways to select exactly 21 objects from a larger set of 42 objects. This set has a specific structure: it contains 21 identical objects and 21 distinct objects.

We need to find the total number of unique combinations of 21 objects that can be formed by choosing from these two groups.

Analyzing the Selection Strategy

To solve this, we can consider how the selection of 21 objects is composed:

• Let $i$ be the number of identical objects we choose.
• Let $d$ be the number of distinct objects we choose.

The total number of objects chosen must be exactly 21. Therefore, the number of identical objects ($i$) and the number of distinct objects ($d$) must satisfy the condition: $$i + d = 21$$

Since there are 21 identical objects available, the possible values for $i$ are integers from 0 to 21 (i.e., $0 \le i \le 21$).

Similarly, since there are 21 distinct objects available, the possible values for $d$ are integers from 0 to 21 (i.e., $0 \le d \le 21$).

Calculating Ways for Identical Objects

For any specific number $i$ of identical objects we decide to choose (where $0 \le i \le 21$), there is only one way to do so, because all these $i$ objects are indistinguishable from each other.

Calculating Ways for Distinct Objects

Once we have decided to choose $i$ identical objects, the constraint $i + d = 21$ dictates that we must choose exactly $d = 21 - i$ distinct objects.

The number of ways to choose $d$ distinct objects from a set of 21 distinct objects is given by the combination formula:

$$ \binom{21}{d} $$

Substituting $d = 21 - i$, the number of ways to choose the distinct objects for a fixed $i$ is:

$$ \binom{21}{21-i} $$

Summing All Possible Combinations

To find the total number of ways to choose 21 objects, we need to sum the possibilities for each potential value of $i$ (the number of identical objects chosen). The value of $i$ can range from 0 to 21.

The total number of ways is:

$$ \text{Total Ways} = \sum_{i=0}^{21} (\text{Ways to choose } i \text{ identical}) \times (\text{Ways to choose } 21-i \text{ distinct}) $$

Plugging in the values we found:

$$ \text{Total Ways} = \sum_{i=0}^{21} 1 \times \binom{21}{21-i} $$

Let's write out the terms in this sum:

• When $i=0$, we choose 0 identical and 21 distinct: $\binom{21}{21-0} = \binom{21}{21}$ ways.
• When $i=1$, we choose 1 identical and 20 distinct: $\binom{21}{21-1} = \binom{21}{20}$ ways.
• When $i=2$, we choose 2 identical and 19 distinct: $\binom{21}{21-2} = \binom{21}{19}$ ways.
• ...
• When $i=21$, we choose 21 identical and 0 distinct: $\binom{21}{21-21} = \binom{21}{0}$ ways.

So, the total sum is:

$$ \text{Total Ways} = \binom{21}{21} + \binom{21}{20} + \binom{21}{19} + \dots + \binom{21}{1} + \binom{21}{0} $$

Applying the Binomial Theorem

This sum is precisely the sum of all binomial coefficients for $n=21$. The binomial theorem states that for any non-negative integer $n$:

$$ \sum_{k=0}^{n} \binom{n}{k} = 2^n $$

In our case, $n=21$. Our sum can be rewritten in the standard order:

$$ \binom{21}{0} + \binom{21}{1} + \dots + \binom{21}{20} + \binom{21}{21} = \sum_{k=0}^{21} \binom{21}{k} $$

According to the binomial theorem, this sum is equal to $2^{21}$.

Final Result

Therefore, the total number of ways of choosing 21 objects out of 42 objects (where 21 are identical and 21 are distinct) is $2^{21}$.