Answer to Question #281738 in Discrete Mathematics for Neha

Question

A person has 8 children of them he takes 3 at a time to a circus. He does not take

the same three children twice to the circus. How many times he will have to go

to circus to ensure that every three children have seen the circus together? In

this case find the number of times a particular child has visited the circus.

From 8 men and 4 women and team of 5 is to be formed. In how many ways can

this be done so as to include at least one woman?

Answer 1

No 3 same children taken again required number = $^{8}C_3=\frac{8!}{3!.5!}=56$

​

Total number of men =8 

Total number of women =4 

Team = 5 persons → at least 1 women 

Case 1→1 women + 4 men ⇒$^4C_1 \times ^{8}C_4=4\times70=280$

Case 2→2 women + 3 men ⇒ $^4C_2 \times ^{8}C_3=6\times56=336$

Case 3→ 3 women + 2 men ⇒$^4C_3 \times ^{8}C_2=4\times28=112$

Case 4→ 4 women + 1 men ⇒$^4C_4 \times ^{8}C_1=1\times8=8$

Total ways $=280+336+112+8=736$