MATH SOLVE

5 months ago

Q:
# Questions (no partial grades if you don't show your work) 1. In a group of 6 boys and 4 girls, four children are to be selected. In how many diffeest weys ces they be selected if at least one boy must be there

Accepted Solution

A:

Answer:Total number of ways will be 209Step-by-step explanation:There are 6 boys and 4 girls in a group and 4 children are to be selected.We have to find the number of ways that 4 children can be selected if at least one boy must be in the group of 4.So the groups can be arranged as(1 Boy + 3 girls), (2 Boy + 2 girls), (3 Boys + 1 girl), (4 boys)Now we will find the combinations in which these arrangements can be done.1 Boy and 3 girls = [tex]^{6}C_{1}\times^{4}C_{3}=6\times4[/tex]=242 Boy and 2 girls=[tex]^{6}C_{2}\times^{4}C_{2}=\frac{6!}{4!\times2!}\times\frac{4!}{2!\times2!}=15\times6=90[/tex]3 Boys and 1 girl = [tex]^{6}C_{3}\times^{4}C_{1}=\frac{6!}{4!\times2!}\times\frac{4!}{3!}=\frac{6\times5\times4}{3 \times2} \times4=80[/tex]4 Boys = [tex]^{6}C_{4}=\frac{6!}{4!\times2!} =\frac{6\times 5}{2\times1}=15[/tex]Now total number of ways = 24 + 90 + 80 + 15 = 209