A sociology seminar has 7 members
(a) They begin each meeting with each member shaking hands with all the other members. No one shakes another member's hand twice. How many handshakes take place?
(b)The professor shakes hands with each member of the seminar and in a different sequence each time. How many sequences of handshakes are possible?
sequences
a) how about C(7,2) = 21
b) You mean only the prof shakes hands?
And he does it again and again?? Silly question, anyway....
In one specific sequence there are 7 handshakes
if the prof shakes hands with each of the 7 members.
Now if we "arrange" that sequence in all its permutations,
the total number possible sequences = 7! = 5040
in a seminar there were twelve people. if each shook their hand all others,how many times were the hands shaken altogether?
(a) To find the number of handshakes that take place in the sociology seminar, we can use the formula for the number of combinations. In this case, we need to choose a pair of two members from the total of 7 members. The formula for combinations is:
nC2 = n! / (2! * (n - 2)!)
where n is the total number of members.
In this case, n = 7:
7C2 = 7! / (2! * (7 - 2)!)
= (7 * 6 * 5 * 4 * 3 * 2 * 1) / (2 * 1 * (5 * 4 * 3 * 2 * 1))
= 21
Therefore, there are 21 handshakes that take place.
(b) To find the number of sequences of handshakes possible when the professor shakes hands with each member of the seminar, we need to calculate the factorial of the total number of members. The formula for factorial is:
n! = n * (n - 1) * (n - 2) * ... * 2 * 1
In this case, n = 7:
7! = 7 * 6 * 5 * 4 * 3 * 2 * 1
= 5,040
Therefore, there are 5,040 sequences of handshakes possible.
To answer both parts of the question, we can consider the concept of combinations and permutations.
(a) In this case, each member needs to shake hands with all the other members. Let's consider each member in turn:
1st member: They will shake hands with 6 other members.
2nd member: They have already shaken hands with the 1st member, so they will shake hands with 5 remaining members.
3rd member: They have already shaken hands with the 1st and 2nd members, so they will shake hands with 4 remaining members.
4th member: They have already shaken hands with the 1st, 2nd, and 3rd members, so they will shake hands with 3 remaining members.
5th member: They have already shaken hands with the 1st, 2nd, 3rd, and 4th members, so they will shake hands with 2 remaining members.
6th member: They have already shaken hands with the 1st, 2nd, 3rd, 4th, and 5th members, so they will shake hands with 1 remaining member.
7th member: They have already shaken hands with the other 6 members.
So, the total number of handshakes will be:
6 + 5 + 4 + 3 + 2 + 1 = 21 handshakes.
(b) For this part, we need to find the total number of possible sequences of handshakes. Since the professor shakes hands with each member of the seminar in a different sequence each time, we need to find the number of permutations.
The number of permutations can be calculated using the factorial function. The factorial of a number n, denoted as n!, is the product of all positive integers less than or equal to n. So, for a seminar with 7 members:
Number of sequences = 7!
Calculating 7!:
7! = 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5040
Hence, there are 5040 possible sequences of handshakes.