Question:
Let 10 Bags \(B_{1},B_{2},\dots,B_{10}\) which contain 21, 22, \dots, 30 different articles respectively. Then the total number of ways to bring out 10 articles from a Bag is:
Let 10 Bags \(B_{1},B_{2},\dots,B_{10}\) which contain 21, 22, \dots, 30 different articles respectively. Then the total number of ways to bring out 10 articles from a Bag is:
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The combination sum identity $\sum_{k=r}^n {}^{k}C_r = {}^{n+1}C_{r+1}$ is often called the Hockey-Stick Identity. Here, since our series starts at ${}^{21}C_{10}$ instead of the baseline ${}^{10}C_{10}$, the missing section can be cleanly accounted for by subtracting the trailing component ${}^{21}C_{11}$!
Updated On: May 28, 2026
- ${}^{31}C_{20}+{}^{21}C_{10}$
- ${}^{31}C_{20}-{}^{21}C_{10}$
- ${}^{30}C_{20}-{}^{20}C_{10}$
- ${}^{30}C_{20}+{}^{20}C_{10}$
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The Correct Option is B
Solution and Explanation
Concept:
The total number of ways to choose a subset of items from independent groups is found by summing their separate combinations. This series can be simplified using Pascal's Identity:
$${}^{n}C_r + {}^{n}C_{r+1} = {}^{n+1}C_{r+1}$$
Step 1: Set up the summation series for the problem.
Let us calculate the number of ways to select 10 articles from each individual bag: $$\text{Ways from } B_1 = {}^{21}C_{10}, \quad \text{Ways from } B_2 = {}^{22}C_{10}, \quad \dots, \quad \text{Ways from } B_{10} = {}^{30}C_{10}$$ Since selecting from each bag represents an independent event choice, sum the combinations together: $$S = {}^{21}C_{10} + {}^{22}C_{10} + {}^{23}C_{10} + \dots + {}^{30}C_{10}$$
Step 2: Apply a helper term to start the telescoping chain.
To use Pascal's identity, we need two terms with matching upper indices. Let us add and subtract a helper combination term ${}^{21}C_{11}$ at the front of the series: $$S = \left( {}^{21}C_{11} + {}^{21}C_{10} \right) + {}^{22}C_{10} + {}^{23}C_{10} + \dots + {}^{30}C_{10} - {}^{21}C_{11}$$
Step 3: Collapse the series step-by-step using Pascal's Identity.
Combine the terms sequentially from left to right:
• First link: ${}^{21}C_{11} + {}^{21}C_{10} = {}^{22}C_{11}$
• Second link: ${}^{22}C_{11} + {}^{22}C_{10} = {}^{23}C_{11}$
• Third link: ${}^{23}C_{11} + {}^{23}C_{10} = {}^{24}C_{11}$ This pattern creates a telescoping chain reaction that rolls through the entire series up to the final term: $${}^{30}C_{11} + {}^{30}C_{10} = {}^{31}C_{11}$$ Bringing back our subtracted helper term leaves: $$S = {}^{31}C_{11} - {}^{21}C_{11} \quad \cdots (1)$$
Step 4: Apply combination symmetry to match the options.
Recall the standard combination reflection identity: ${}^{n}C_r = {}^{n}C_{n-r}$. Let us modify both terms in equation (1):
• ${}^{31}C_{11} = {}^{31}C_{31-11} = {}^{31}C_{20}$
• ${}^{21}C_{11} = {}^{21}C_{21-11} = {}^{21}C_{10}$ Substituting these symmetric terms gives our final simplified answer: $$S = {}^{31}C_{20} - {}^{21}C_{10}$$ This matches option (B) perfectly.
Let us calculate the number of ways to select 10 articles from each individual bag: $$\text{Ways from } B_1 = {}^{21}C_{10}, \quad \text{Ways from } B_2 = {}^{22}C_{10}, \quad \dots, \quad \text{Ways from } B_{10} = {}^{30}C_{10}$$ Since selecting from each bag represents an independent event choice, sum the combinations together: $$S = {}^{21}C_{10} + {}^{22}C_{10} + {}^{23}C_{10} + \dots + {}^{30}C_{10}$$
Step 2: Apply a helper term to start the telescoping chain.
To use Pascal's identity, we need two terms with matching upper indices. Let us add and subtract a helper combination term ${}^{21}C_{11}$ at the front of the series: $$S = \left( {}^{21}C_{11} + {}^{21}C_{10} \right) + {}^{22}C_{10} + {}^{23}C_{10} + \dots + {}^{30}C_{10} - {}^{21}C_{11}$$
Step 3: Collapse the series step-by-step using Pascal's Identity.
Combine the terms sequentially from left to right:
• First link: ${}^{21}C_{11} + {}^{21}C_{10} = {}^{22}C_{11}$
• Second link: ${}^{22}C_{11} + {}^{22}C_{10} = {}^{23}C_{11}$
• Third link: ${}^{23}C_{11} + {}^{23}C_{10} = {}^{24}C_{11}$ This pattern creates a telescoping chain reaction that rolls through the entire series up to the final term: $${}^{30}C_{11} + {}^{30}C_{10} = {}^{31}C_{11}$$ Bringing back our subtracted helper term leaves: $$S = {}^{31}C_{11} - {}^{21}C_{11} \quad \cdots (1)$$
Step 4: Apply combination symmetry to match the options.
Recall the standard combination reflection identity: ${}^{n}C_r = {}^{n}C_{n-r}$. Let us modify both terms in equation (1):
• ${}^{31}C_{11} = {}^{31}C_{31-11} = {}^{31}C_{20}$
• ${}^{21}C_{11} = {}^{21}C_{21-11} = {}^{21}C_{10}$ Substituting these symmetric terms gives our final simplified answer: $$S = {}^{31}C_{20} - {}^{21}C_{10}$$ This matches option (B) perfectly.
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