Question

At a party there are 20 men and 10 women. Each man shakes hand only with all the men and each woman shakes hand only with all women. Find the total number of hand shakes at the party.​

Answer 1

Answer:

500

Step-by-step explanation:

(20×20)+10×10

= 400+(10×10)

= 400+100

= 500

Hope it helped.

Answer 2

Answer:

Total hand shakes at the party will be 235

Step-by-step explanation:

In this Question.

We have been given that there are 20 men and 10 women in the party and every men shakes hand only with men only and women shakes hand with women only.

We have to find the total number of handshakes in the party

As we know if 1 men shakes hand with the remaining men i.e 19 men,Then total number of handshakes will be 19.Handshakes should not be Repeated.2nd men will shake hand with the remaining men i.e 18 men.

So the total number of handshakes of men will be sum of (0+1+2+......+19)

We know the sum of first n numbers $=\frac{n(n+1)}{2}$

Therefore sum of first 19 numbers will be $=\frac{19(19+1)}{2}$

                                                                     $=\frac{19\times 20}{2}$

                                                                     $=19\times 10$

                                                                            =190

As we know if 1 women shakes hand with the remaining men i.e 9 women,Then total number of handshakes will be 9 and we should note that Handshakes should not be Repeated.2nd women will shake hand with the remaining 8women i.e 8 women.

So Total number of handshakes of women is the sum of(0+1+2+.....+9)

Therefore sum of first 9 numbers will be $=\frac{9(9+1)}{2}$

                                                                   $=\frac{9\times 10}{2}$

                                                                   $=9\times 5$

                                                                          =45

Therefore total number of handshakes is = 190+45

                                                                     =235