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Animated Solution for Mathematics - Permutations and Combinations: A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is

Select Answer:

Progressive Hints

Hint 1

Hint 1: Visualize the exam paper as two distinct buckets: the first 5 questions and the remaining 8 questions. The constraint 'at least 4 from the first 5' is a fork in the road. You must explore every path that satisfies this condition.

Hint 2

Hint 2: Since the student must choose exactly 10 questions in total, consider the two mutually exclusive scenarios: choosing exactly 4 from the first bucket (which forces a specific number from the second) and choosing exactly 5 from the first bucket (which forces another specific number from the second).

Hint 3

Hint 3: Use the combination formula

^{n} C_{r} = \frac{n !}{r ! ( n - r )!}

to calculate the ways for each scenario. Your starting equation for the total is: Total = (

^{5} C_{4} \times^{8} C_{6}

) + (

^{5} C_{4} \times^{8} C_{5}

)? No, wait, check the second case: (

^{5} C_{5} \times^{8} C_{5}

Visualized Solution

Visualizing the Groups

Total Questions: $13$
Group 1: First $5$ questions
Group 2: Remaining $13 - 5 = 8$ questions
Goal: Select $10$ questions in total.

Analyzing the 'At Least' Constraint

Constraint: At least $4$ from Group 1.
This implies two mutually exclusive cases:
Case 1: Exactly $4$ from Group 1 AND $6$ from Group 2.
Case 2: Exactly $5$ from Group 1 AND $5$ from Group 2.

Case 1: Selecting Exactly $4$ from Group 1

Case 1: $4$ from Group 1 and $6$ from Group 2.
Ways to choose from Group 1: $^{5} C_{4}$
Ways to choose from Group 2: $^{8} C_{6}$
Total ways for Case 1: $^{5} C_{4} \times^{8} C_{6}$

Calculating Case 1

$^{5} C_{4} =^{5} C_{1} = 5$
$^{8} C_{6} =^{8} C_{2} = \frac{8 \times 7}{2 \times 1} = 28$
Total for Case 1: $5 \times 28 = 140$

Case 2: Selecting Exactly $5$ from Group 1

Case 2: $5$ from Group 1 and $5$ from Group 2.
Ways to choose from Group 1: $^{5} C_{5}$
Ways to choose from Group 2: $^{8} C_{5}$
Total ways for Case 2: $^{5} C_{5} \times^{8} C_{5}$

Calculating Case 2

$^{5} C_{5} = 1$
$^{8} C_{5} =^{8} C_{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$
Total for Case 2: $1 \times 56 = 56$

Final Summation

Total choices = (Ways in Case 1) + (Ways in Case 2)
Total choices = $140 + 56$
Total choices = $196$

The Way Forward

Key Takeaway: Break 'at least' constraints into mutually exclusive cases.
Pro Tip: Use $^{n} C_{r} =^{n} C_{n - r}$ to simplify calculations.

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The Sigma Insight: Combinations and Selection

Solution Diagram

Analyzing the Setup

We are tasked with selecting

10

questions out of a total of

13

. The questions are divided into two distinct groups: Group 1: The first $5$ questions. Group 2: The remaining

8

questions.

The constraint requires us to select at least $4$ questions from Group 1. This implies two mutually exclusive scenarios: selecting exactly

4

questions from Group 1, or selecting all

5

questions from Group 1.

The First Path

Case 1

In this scenario, we choose exactly

4

questions from the first

5

and the remaining

6

questions from the

8

available in Group 2. The number of ways to perform this selection is given by the product of the combinations:

N_{1} =^{5} C_{4} \times^{8} C_{6}

We know that

^{5} C_{4} =^{5} C_{1} = 5

. Furthermore, using the symmetry property

^{n} C_{r} =^{n} C_{n - r}

, we calculate

^{8} C_{6}

as:

^{8} C_{6} =^{8} C_{2} = \frac{8 \times 7}{2 \times 1} = 28

Multiplying these values, we find the total ways for Case 1:

N_{1} = 5 \times 28 = 140

The Second Path

Case 2

In this scenario, we choose all

5

questions from Group 1. To reach our total of

10

questions, we must select the remaining

5

questions from the

8

available in Group 2.

N_{2} =^{5} C_{5} \times^{8} C_{5}

Since

^{5} C_{5} = 1

and

^{8} C_{5} =^{8} C_{3}

, we calculate the latter as:

^{8} C_{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56

Multiplying these values, we find the total ways for Case 2:

N_{2} = 1 \times 56 = 56

The Synthesis

Because these two cases are mutually exclusive, we apply the Addition Principle to find the total number of valid selections.

Total Ways = N_{1} + N_{2}

Total Ways = 140 + 56 = 196

Final Result: The total number of ways to select the questions under the given constraints is $196$ .

Common Traps & FAQs

Wait, why do we split this into two cases? Can't we just pick 4 questions first and then pick 6 more from the remaining 9?

That is a classic trap! If you pick 4 from the first 5 and then 6 from the remaining 9, you are overcounting. Specifically, you might accidentally pick the 5th question from the first group as part of your 'remaining 9' selection, which would actually mean you picked 5 from the first group. By splitting into mutually exclusive cases (exactly 4 vs exactly 5), we ensure every unique combination is counted exactly once without overlap.

Why do we add the results of the two cases instead of multiplying them?

Great question! We use the Fundamental Principle of Addition because these cases are mutually exclusive. The student either chooses exactly 4 from the first group OR exactly 5. They cannot do both simultaneously. When scenarios are 'OR' conditions, we add. When they are 'AND' conditions (like picking questions from group 1 AND group 2), we multiply.

Is there a faster way to do this using the complement method?

You could, but it might be more complex here. The complement would be 'Total ways to pick 10' minus 'Ways to pick fewer than 4 from the first group'. 'Fewer than 4' means 0, 1, 2, or 3 from the first group. That is four cases to calculate! Our current method of just two cases is much more efficient. Always look for the path of least resistance!

Animated Solution for Mathematics - Permutations and Combinations: A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is

Select Answer:

Visualized Solution

Visualizing the Groups

Analyzing the 'At Least' Constraint

Case 1: Selecting Exactly $4$ from Group 1

Calculating Case 1

Case 2: Selecting Exactly $5$ from Group 1

Calculating Case 2

Final Summation

The Way Forward

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The First Path

The Second Path

The Synthesis

Similar Questions

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Animated Solution for Mathematics - Permutations and Combinations: A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is

Select Answer:

Visualized Solution

Visualizing the Groups

Analyzing the 'At Least' Constraint

Case 1: Selecting Exactly .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 4 from Group 1

Calculating Case 1

Case 2: Selecting Exactly .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 5 from Group 1

Calculating Case 2

Final Summation

The Way Forward

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The First Path

The Second Path

The Synthesis

Similar Questions

A test consists of 6 multiple choice questions, each having 4 alternative answers of which only one is correct. The number of ways, in which a candidate answers all six questions such that exactly four of the answers are correct, is

A boy needs to select five courses from 12 available courses, out of which 5 courses are language courses. If he can choose at most two language courses, then the number of ways he can choose five courses is ____.

There are 3 sections in a question paper and each section contains 5 questions. A candidate has to answer a total of 5 questions, choosing at least one question from each section. Then the number of ways, in which the candidate can choose the questions, is :

Consider 4 boxes, where each box contains 3 red balls and 2 blue balls. Assume that all 20 balls are distinct. In how many different ways can 10 balls be chosen from these 4 boxes so that from each box at least one red ball and one blue ball are chosen?

At an election, a voter may vote for any number of candidates, not greater than the number to be elected. There are 10 candidates and 4 are of be selected, if a voter votes for at least one candidate, then the number of ways in which he can vote is

A debate club consists of 6 girls and 4 boys. A team of 4 members is to be selected from this club including the selection of a captain (from among these 4 members) for the team. If the team has to include at most one boy, then the number of ways of selecting the team is

The number of ways of getting a sum 16 on throwing a dice four times is______

Case 1: Selecting Exactly $4$ from Group 1

Case 2: Selecting Exactly $5$ from Group 1