In a pack of 42 cards, 3 cards are chosen one after the other. Find the number of ways this can be done without replacement:
Understanding the Card Selection Problem
This problem asks for the total number of different sequences in which we can select 3 cards from a set of 42 cards, with the condition that once a card is chosen, it is not put back into the pack (this is what "without replacement" means).
Problem Breakdown
- Total number of cards (n): 42
- Number of cards to choose (k): 3
- Selection process: One after the other (order matters)
- Condition: Without replacement (no repetition)
Step-by-Step Solution
This scenario involves permutations because the order in which the cards are chosen matters. We need to find the number of permutations of 42 items taken 3 at a time, denoted as $P(42, 3)$ or $_{42}P_3$.
The calculation can be done step-by-step:
- Choosing the first card: You have 42 distinct cards to choose from. So, there are 42 possibilities for the first pick.
- Choosing the second card: After picking one card and not replacing it, you now have 41 cards left in the pack. So, there are 41 possibilities for the second pick.
- Choosing the third card: After picking two cards, you have 40 cards remaining. So, there are 40 possibilities for the third pick.
To find the total number of ways to choose these 3 cards in sequence, we multiply the number of possibilities at each step:
Number of ways = (Choices for 1st card) $\times$ (Choices for 2nd card) $\times$ (Choices for 3rd card)
Number of ways = $42 \times 41 \times 40$
Calculation
Let's perform the multiplication:
- $42 \times 41 = 1722$
- $1722 \times 40 = 68880$
So, there are 68,880 different ways to choose 3 cards one after the other from a pack of 42 cards without replacement.
Permutation Formula
Alternatively, we can use the permutation formula:
$P(n, k) = \frac{n!}{(n-k)!}$Here, $n=42$ and $k=3$.
$P(42, 3) = \frac{42!}{(42-3)!} = \frac{42!}{39!}$ $P(42, 3) = \frac{42 \times 41 \times 40 \times 39 \times 38 \times \dots \times 1}{39 \times 38 \times \dots \times 1}$The terms from $39!$ cancel out, leaving:
$P(42, 3) = 42 \times 41 \times 40 = 68880$Conclusion
The total number of ways to choose 3 cards from a pack of 42 cards, one after the other without replacement, is 68,880.
Important Questions from Permutation and Combination
m parallel lines cut n parallel lines giving rise to 60 parallelograms. What is the value of (m + n) ?
5-digit numbers are formed using the digits 0, 1, 2, 4, 5 without repetition. What is the percentage of numbers which are greater than 50,000 ?
In a race, there are 4 members in a team. Each member has to cover 5 km one after another. If the total time taken is 30 minutes, then what would have been the average speed?
If Quantity A is the number of ways to assign a number from 1 to 5 without repetition to each of four people, and Quantity B is the number of ways to assign a number from 1 to 5 without repetition to each of 5 people, then which of the following statements is correct with respect to Quantities A and B?
Which of the following muscles regulates the exit of food from the stomach into the small intestine?
