A disc jockey has to choose three songs for the last few minutes of his evening showIf there are nine songs that he feels are appropriate for that time slot, then how many
ways can he choose and arrange to play three of those nine songs?

I'll do the first one for you. it isn't very hard at all.on number one, you have (50 × 3), which equals 150, and (3 × 3), which equals 9. after you establish that, you add 150 and 9 together. so, (150 + 9) equals 159.this means that 53 × 3 = 159.

Sin(a)+cos(a)+cot(a)-1=2(csc(x)+1)