Problem Analysis: Total Handshakes
This problem involves calculating the total number of unique pairs that can be formed from a group of 40 individuals, where each pair represents a handshake. Since the order of people in a handshake does not matter (Person A shaking Person B's hand is the same handshake as Person B shaking Person A's hand), this is a combination problem.
Applying the Combination Formula
The number of combinations of choosing k items from a set of n items is given by the formula:
$C(n, k) = \frac{n!}{k!(n-k)!}$
For the handshake problem, we are choosing groups of 2 people (k=2) from a total of 40 people (n=40). The formula simplifies to:
$C(n, 2) = \frac{n(n-1)}{2}$
Calculation for 40 Persons
Given n = 40 persons:
- Substitute the value of n into the simplified formula:
$C(40, 2) = \frac{40 \times (40 - 1)}{2}$
- Perform the subtraction:
$C(40, 2) = \frac{40 \times 39}{2}$
- Simplify the expression:
$C(40, 2) = \frac{1560}{2}$
- Calculate the final result:
$C(40, 2) = 780$
Therefore, the total number of handshakes will be 780.
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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?
