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Q6:
Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.
Solution :
Step 1: Identifying the Mathematical Framework
To determine the number of combinations, we employ the combinatorial formula $^nC_r = \frac{n!}{r!(n-r)!}$, where $n$ is the total number of items available and $r$ is the number of items to be selected [Fundamental Counting Principle & Definition of Combinations].
Step 2: Partitioning the Set of Cards
A standard deck of 52 cards is partitioned into two distinct categories based on the problem constraints:
- Category A (Aces): There are $4$ aces in a standard deck.
- Category B (Non-Aces): The remaining cards in the deck are $52 - 4 = 48$ cards.
Step 3: Calculating Independent Selections
The requirement is a 5-card combination containing exactly one ace. This implies the selection process consists of two simultaneous events:
- Selecting 1 Ace: We must choose 1 card from the 4 available aces.
Calculation: $^4C_1 = \frac{4!}{1!(4-1)!} = 4$ [Property: $^nC_1 = n$] - Selecting the Remaining 4 Cards: Since the total hand must contain 5 cards and 1 is already an ace, we must choose 4 additional cards from the 48 non-ace cards.
Calculation: $^{48}C_4 = \frac{48!}{4!(48-4)!} = \frac{48 \times 47 \times 46 \times 45}{4 \times 3 \times 2 \times 1}$
Step 4: Executing Numerical Computation
Applying the division for the non-ace selection:
- $^{48}C_4 = \frac{4,669,920}{24} = 194,580$
By the Multiplication Principle of counting, the total number of combinations is the product of the two independent selections [Fundamental Principle of Counting]:
Total Combinations $= (^4C_1) \times (^{48}C_4)$
Total Combinations $= 4 \times 194,580$
Total Combinations $= 778,320$
Final Solution: The total number of 5-card combinations containing exactly one ace is 778,320.
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.
- 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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