For this problem, assume 7 males audition, one of them being George, 6 females audition, one of them being Margaret, and 4 children audition. The casting director has 3 male roles available, 1 female role available, and 2 child roles available. (1) How many different ways can these roles be filled from these auditioners? (2) How many different ways can these roles be filled if exactly one of George and Margaret gets a part? (3) What is the probability (if the roles are filled at random) of both George and Margaret getting a part?

Answer:1.12962.5703.[tex]\frac{6}{91}[/tex]Step-by-step explanation:We are given that Number of males=7 including GeorgeNumber of females=6 including Margaret Number of children=4Number of male selecting for roles=3Number of females selecting for roles=1Number of child selecting for roles=21.We have to find the number of ways can these roles be filled from these auditioners.Total number of ways=[tex]7C_2\times 6C_1\times 4C_2[/tex][tex]nC_r=\frac{n!}{r!(n-r)!}[/tex] Using this formula Total number of ways=[tex]\frac{9\times 8\times 7!}{2!7!}\times \frac{6\times 5!}{1!5!}\times \frac{4\times 3\times 2!}{2!\times 2\times 1}[/tex]Total number of ways=12962.We have to find number of ways can these roles be filled if exactly one of George and Margaret gets a part.If George gets a part then Margaret outTotal number of ways=[tex]6C_2\times 5C_1\times 4C_2=\frac{6\times 5\times 4!}{2\times 1\cdot4!}\times 5\times \frac{4\times 3\times 2!}{2\times 1\cdot 2!}=450[/tex]If Margaret gets a part then George outNumber of ways=[tex]6C_3\times 4C_2=\frac{6\times 5\times 4\times 3!}{3\times 2\times 1\times 3!}\times \frac{4\times 3\times 2}{2!\times 2\times 1}=120[/tex]Therefore, total number of ways can these roles be filled if exactly one of George and Margaret gets a part=450+120=5703.We have to find the probability of both George and Margaret getting  a part.Total number of audition=7+6+4=17Except George and Margaret , number of auditions=15Number of males=6Probability=[tex]\frac{number\;of\;favorable\;cases}{total\;number\;cases}[/tex]The probability of both George and Margaret getting a  part=[tex]\frac{6C_2\times 4C_2}{15C_4}=\frac{\frac{6!}{2!4!}\times \frac{4!}{2!2!}}{\frac{15!}{4!11!}}[/tex]The probability of both George and Margaret getting a  part=[tex]\frac{6\times 5\times 4!}{2\times 1\times 4!}\times \frac{4\times 3\times 2!}{2!\times 2\times 1}\times \frac{4\times 3\times 2\times 11!}{15\times 14\times 13\times 12\times 11!}[/tex]The probability of both George and Margaret getting a  part=[tex]\frac{6}{91}[/tex]