Five people meet and exchange handshakes. If exactly 3 of the people

Question

Five people meet and exchange handshakes. If exactly 3 of the people each shake hands with 3 people, and if 1 person shakes hands with only 1 other person, then what is the least number of handshakes that could have been exchanged?

A. 9
B. 7
C. 6
D. 5
D. 3

Show Answer

Official Answer

C

A. 9
B. 7
C. 6
D. 5
D. 3

Nidzo · Answer

Let the five people be  A, B, C, D & E

Let  A, B, C & D  shake one another's hands. This will result in  \frac{4!}{2!2!} = 6  handshakes occurring.

This means that 4 people ( A, B, C & D ) have all shaken 3 hands each. We are told that exactly 3 people shake 3 hands and that 1 person shakes only 1 hand. We already have 4 that have shaken 3 hands, which is one too many. So we let one of the 4 people who have shaken 3 hands shake 1 more. Let  D  shake hands with  E , which means that in total 7 handshakes have happened.

Now we have:
 A, B & C:  with 3 handshakes each (as stated in the question). 
 E:  shakes 1 person's hand (as stated in the question). 
 D:  shakes 4 people's hands.

Answer B.

poorvi125 · Answer

Let there be five people - A, B, C, D and E.
Let A, B and C be the people who have three handshakes each. I.e., A shakes with B, C and D; B shakes with A, C and E; and C shakes with A, B and E. So, handshakes are as follows:

A - B
A - C
A - D
B - C
B - E

Now, let D be the person who shakes with only one person. As we have already accounted for D's handshake in above list, we will not double count it. Now, since E can not be the one who has either three handshaeks or one handshake, the minimum no. of handshakes he much have is two. As we have already accounted for his handshake with B above, he will have one more handshake as below:

E - C

Total no. of handshakes is, therefore, 6.

Answer - C.

Gemmie · Answer

Let's denote the 5 people as A, B, C, D, E.

Initial Observations:

If 3 people each shake hands with 3 others, then these 3 people have collectively engaged in 3×3=9 handshake events.

However, each handshake involves two people, so the total for these 3 people is counted twice, hence they actually account for  9 \div 2 = 4.5  handshakes among themselves.

Since handshakes must be integers, there is an inherent symmetry and some overlap with the handshakes involving other people.

Handshake Graph Construction:

Let A, B, C be the 3 people who each shake hands with 3 others.
Let D be the person who shakes hands with only 1 person.
E must be the remaining person.

Forming Connections:

A,B,C need to shake hands with each other, so each has 2 handshakes within the group.

Each of A, B, C must also shake hands with another person outside their group to satisfy the condition of shaking hands with exactly 3 people. This implies:

+) A shakes hands with B, C, and either D or E.
+) B shakes hands with A, C, and either D or E.
+) C shakes hands with A, B, and either D or E.Without loss of generality, assume:

=> A shakes hands with D.
B shakes hands with E.
C shakes hands with E.
(so that D only shakes hand with only 1 person)

Counting the Handshakes:

+) Handshakes among A,B,C: (A−B), (A−C), (B−C). These are 3 handshakes.
+) Additional handshakes to meet the condition: (A-D), (B-E), (C-E) => 3 additional handshakes.Thus, the total handshakes count is 6

chalant · Answer

3x2=6, as in permutations among the 3 people shaking hands with 3 other people.

The statement that 1 person shook hands with only one other person doesn't matter here because it's up to interpretation whether that's already included in 3x2. We're aiming to minimize the number of handshakes and therefore assume that it's indeed contained within 3x2=6.

Posted from my mobile device

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Five people meet and exchange handshakes. If exactly 3 of the people

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Five people meet and exchange handshakes. If exactly 3 of the people

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