Shaking hands at Aakash's party

We use complementary counting. First we count the number of handshakes between 20 people. This is $\dbinom{20}{2}=190$ Now, we find the number of handshakes which do not occur. This is simply 10, one for each married couple. The answer is $190-10=\boxed{180}$ .

There are ten couples so that there are 20 people.Each of those 20 would shake hands with 18 others.
Hence, we have $20 \times 18$
However, this would count the handshakes between each pair twice.
So, we must divide by $2$ .
Thus, $(20 \times 18)/2$ = $\boxed{180}$
If we generalize this problem, that is if we have n people. Let's determine then the number of different handshakes:

The formula is : n(n-1)/2

There are $10\times2=20$ people at the party. Each person shakes hands with everyone besides him/herself and his/her spouse, so each person shakes the hand of $18$ people. Since there are 20 people people and each shakes the hands of 18 people, there are a total of $\frac{18\times20}{2}$ handshakes; one handshake occurs with two people so we must divide by 2. Computing, we find that the answer is equal to $\boxed{180}$ .

10c2-10=180

Total number of handshakes (including that of the couples ) = 20 C 2 = $\frac{20!}{18!.2!}$ = 190

Now , number of handshakes between couples = 10 as 10 couples are present .

Therefore, total number of handshakes (excluding couple-handshakes) = 190 - 10 = $\boxed{180}$

see every possible handshake so =20c2 (selecting any two gives a handshake)-10=180

First there are handshakes between only men these are 9+8+7+6+5+4+3+2+1= 45 Similarly only between women there will be 45 handshakes. Between men and women, there will be 9*10 = 90 handshakes. A total of 45+45+90 = 180 handshakes.

Let A and B be one of the married couples. A will shake hands with 18 people(other than himself and his wife B), similarly B will shake hands with 18 people.But the next couple will shake hands with 16 people and so on.

Total handshakes = $(18+18)+(16+16)+(14+14)...........+(0+0) = \boxed{180}$

NICE problem

There are a total of 20 people, as each man/woman cannot shake hand with their spouse or themselves, that means each person can shake had with 18 people. The first person's spouse can also shake hand with 18 people because his/her spouse had not shaken hand with him/her. The third and fourth person can only do so with 16 people, because they can shake hands with anyone bar themselves, their spouse, and the first couple i.e. 20-4=16. The pattern goes on for the next couple etc. so the no. of hand shakes is:

18+18+16+16+14+14+12+12+10+10+8+8+6+6+4+4+2+2

or 2(18+16+14+12+10+8+6+4+2)

Do you see the pattern now?

first 10 couples r there so total 20 no. of people then when they'll shake hands except their spouse so.....shaking would be with 18 people 18*10=180

total handshaskes=360 Except for his or her own spouse=2

Then the total handshakes in party=360/2=180

There are $10 \times 2=20$ people at the party.

There are $\frac{19 \times 20}{2}=190$ ways to shake hands with each other.

However, one cannot shake hands with his own spouse, and there are 10 ways to do this.

Hence, the total number of handshakes is $190-10=180$ .

The solution to the above problem can be thought in 2 ways ....

1) 20C2 - 10 {i.e. no. of ways in which 2 person can be selected to perform handshake , but we have to subtract those cases in which a person is shaking hand with his/her spouse i.e 10}

2) There are 10 couples , therefore pick any couple (9 would be left) therefore no. of handshakes by this couple with other's can be 9x2x2 , similarlly for second couple handshakes would be 8x2x2 ..... thus total no. of handshakes could be 9x2x2 + 8x2x2 + 7x2x2 + ...... + 1x2x2 = 180

The total number of handshakes shared between 20 people can be counted as such :

20!/(20-2)!(2!) = 190.

Since this includes the 10 handshakes shared between the spouses, there are 190-10 = 180 handshakes.

10 couples: first case consider the total number of handshakes (involving men) to do this consider the couples arranged in a row...now the man in the first couple gets to shake hands with (9x2) people....the man of the second couple shakes hands with( 8x2) ...third man (7x2)......this goes on till (2x1)....sum the series=90. Now considering women its the same result. Hence total=(90x2)=180

Person A1 has 18 hands to shake. Person A2 (the spouse) also 18 hands to shake. However, Person B1 has 16. Person B2 has 16. If you write it all out it will be 18+18+16+16+14+14+...+2+2, which we can write as 4(9+8+7+6+...+1). That is equal to 4(45), which is 180

There are a total of 20 persons. Now, suppose all couples are made to stand in a line. 1st husband comes and shakes hand with all except himself and his wife i.e he does 18 handshakes..Now, he goes away..His wife shakes hand with all 18 persons remaining...Hence, it would be 18+18+16+16+14+14+12+12+10+10................+2+2 = 180

Each of 20 guests shakes 18 hands (20 guests, minus themselves and their spouse). This is a total of 360 hands.

However, this counts each handshake twice: A has shaken B's hand, and B has shaken A's hand, but this is counted as two shakes when in reality it is only one. So, divide the number of hands by 2 to get the number of handshakes, 180 .

There are 20 people in total. Because each person does not shake hands with themselves or their spouse, each person will shake hands with 18 other people. We can view 10 husbands (or 10 wives) shaking hands with the other 9 wives and 9 husbands. Therefore, the total handshakes = * 18 10 = 180 **

Since there are 20 people (10 x 2 people) in the crowd and they shake hands with everyone except their spouses, we can assume that that for every one (20 people), he or she will shake hands with 9 couples giving the equation of 9 x 20 =180.

Since there are 10 couples, there must be 20 people at the party. Assume the restriction that each person doesn't shake their spouse's hand doesn't exist. In this case there would be (20*19)2 handshakes. However, we must subtract 10 from that since each person does not shake their spouse's hand giving 190-10=180 handshakes.