There are 18 students involved in the class production of arsenic and old lace. a) in how many ways can the teacher cast the play if there are five male roles and seven female roles and the class has nine male and nine female students? b) in how many ways can the teacher cast the play if jean will play the young female part only if jovane plays the male lead? c) in how many ways can the teacher cast the play if all the roles could be played by either a male or a female student?

a. 9Permutation5 × 9Permutation7

= 2743372800

b. for boys there are 9Combination5 = 126 combos

of which jovane

and in 14 of which 1/5 of 70 he will have the lead

for girls there are 9C7 = 36 combos

in 28 of which 7/9 of 36 jean will figure

there are thus a total of 126×36 combos

and we multiplied them by 5!×7!

to get permutations needed for cases:

- jovane chosen & not given lead & jean chosen : 56×28

- jovane not chosen & jean chosen: 56×28

total such combos = 3136

we have now to multiply by 5!×6×6!

to get permutations with restriction,

For instance(5!×7! - 5!×6×6!) = 86400 less for each such combo

so finally, all we have to do

is to subtract 3136×86400 from ans of qa to give

2743372800 - 270950400

= 2472422400

c. 18Permutation12

= 8892185702400

it's d.

step-by-step explanation:

because: -1(4x + 1y) +2 (1x-3y) distribute each number into the parenthesis.

then you would get: -4x - 1y + 2x -6y combine like terms

and get: 2x -7y

hope my answer has you if not then i'm sorry.