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
On a chess board, in how many different ways can 6 consecutive squares be chosen on the diagonals along a straight path ?
There are 6 persons arranged in a row. Another person has to shake hands with 3 of them so that he should not shake hands with two consecutive persons. In how many distinct possible combinations can the handshakes take place ?
In a tournament of Chess having 150 entrants, a player is eliminated whenever he loses a match. It is given that no match results in a tie/draw. How many matches are played in the entire tournament?
The letters A, B, C, D and E are arranged in such a way that there are exactly two letters between A and E. How many such arrangements are possible?
There is a numeric lock which has a 3-digit PIN. The PIN contains digits 1 to 7. There is no repetition of digits. The digits in the PIN from left to right are in decreasing order. Any two digits in the PIN differ by at least 2. How many maximum attempts does one need to find out the PIN with certainty?
