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Q7:
In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers?
Solution :
Step 1: Identifying the Mathematical Framework
To determine the number of ways to form the team, we apply the Fundamental Counting Principle and the Combinations formula. Since the order of selection within the team does not matter, we use the formula for combinations denoted as $^nC_r = \frac{n!}{r!(n-r)!}$. [Per the Principle of Combinations]
Step 2: Categorization of Players
The total pool of 17 players is partitioned into two distinct groups based on the capability to bowl:
- Group A (Bowlers): $n_1 = 5$ players.
- Group B (Non-bowlers): $n_2 = 17 - 5 = 12$ players.
Step 3: Defining Selection Constraints
A team of 11 players must be formed with exactly 4 bowlers. This implies:
- Selection from Group A (Bowlers): Select $r_1 = 4$ out of $n_1 = 5$.
- Selection from Group B (Non-bowlers): Since the total team size is 11, the remaining spots $r_2 = 11 - 4 = 7$ must be filled from the $n_2 = 12$ non-bowlers.
Step 4: Calculating Individual Combinations
Selection of Bowlers ($^5C_4$):
$^5C_4 = \frac{5!}{4!(5-4)!} = \frac{5 \times 4!}{4! \times 1!} = 5$
Selection of Non-bowlers ($^{12}C_7$):
$^{12}C_7 = \frac{12!}{7!(12-7)!} = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7!}{7! \times 5!}$
$^{12}C_7 = \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1} = \frac{95040}{120} = 792$
Step 5: Applying the Multiplication Principle
According to the Multiplication Principle, if one event can occur in $m$ ways and a second independent event can occur in $n$ ways, the total number of ways the events can occur in sequence is $m \times n$. [Per the Fundamental Counting Principle]
Total ways = $^5C_4 \times ^{12}C_7$
Total ways = $5 \times 792 = 3960$
Visual Representation of Combinatorial Selection
Final Solution: The total number of ways to select the team is 3960.
More Questions from Class 11 Mathematics Permutations and Combinations EXERCISE 6.4
- Q1: If ${}^nC_8 = {}^nC_2$, find ${}^nC_2$.
- Q2(i): Determine $n$ if (i) ${}^{2n}C_3 : {}^nC_3 = 12 : 1$
- Q2(ii): Determine $n$ if (ii) ${}^{2n}C_3 : {}^nC_3 = 11 : 1$
- Q3: How many chords can be drawn through 21 points on a circle?
- Q4: In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?
- Q5: Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.
- Q6: Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.
- Q8: A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.
- Q9: In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?
CBSE Solutions for Class 11 Mathematics Permutations and Combinations
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