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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.
Solution :
Step 1: Analyzing the Selection Criteria
The problem requires the selection of a total of $9$ balls from three distinct groups of colored balls. The constraints are defined as follows:
- Total red balls available ($n_R$) = $6$
- Total white balls available ($n_W$) = $5$
- Total blue balls available ($n_B$) = $5$
- Selection requirement: Exactly $3$ balls of each color.
Step 2: Mathematical Formulation using Combinations
Since the order of selection does not matter within the color groups, we employ the combination formula, denoted as $^nC_r = \frac{n!}{r!(n-r)!}$, where $n$ is the total number of items to choose from and $r$ is the number of items chosen [Per the Principle of Combinatorics].
Step 3: Calculating Individual Selections
We calculate the number of ways to select 3 balls for each color independently:
(i) Selection of 3 red balls from 6:
$$^6C_3 = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$$
(ii) Selection of 3 white balls from 5:
$$^5C_3 = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10$$
(iii) Selection of 3 blue balls from 5:
$$^5C_3 = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10$$
Step 4: Applying the Fundamental Counting Principle
According to the Fundamental Counting Principle (Multiplication Principle), if one task can be done in $m$ ways and a second task can be done in $n$ ways, then both tasks can be performed in $m \times n$ ways. Since the selections of red, white, and blue balls are independent operations, we multiply the individual results [Per the Fundamental Principle of Counting].
Total Ways = $^6C_3 \times ^5C_3 \times ^5C_3$
Total Ways = $20 \times 10 \times 10$
Final Solution: 2000 ways
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?
- Q6: Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.
- 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?
- 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
Chapters in CBSE - Class 11 Mathematics
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