How many ways can 5 sopranos and 4 altos be selected from 7 sopranos and 9 altos?

Answer 1

First, let us consider each case separately.

Looking at the first case, we need to find how many ways we can choose 5 sopranos from 7 sopranos. When we need to find the amount of combinations of n objects when we can only choose r objects, we can use the formula:

combinations = n!/r!(n-r)!

The ! symbol means to multiply that number by all the numbers before it.

So 4! = 4 * 3 * 2 * 1 = 24

n = the total number of sopranos we can choose from (7 sopranos)

r = the amount of sopranos we can actually choose (5 sopranos)

So plugging everything into the formula:

combinations = 7!/5!(7-5)! = 7!/5!*2! = (7*6*5*4*3*2)/(5*4*3*2)*(2) = 21 different combinations.

Next, let's consider the altos.

Our formula (combinations = n!/r!(n-r)!) is the same so we just need to plug in a different set of values.

n = the total number of altos we can choose from (9 altos)

r = the amount of altos we can actually choose (4 altos)

So, using the formula:

combinations = 9!/4!(9-4)! = 9!/4!*5! = (9*8*7*6*5*4*3*2)/(5*4*3*2)*(4*3*2) = 126 different combinations.

If we treat the choosing of the alto's and soprano's as two independent events, then we can figure out the number of combinations of both events by multiplying the number of combinations for both. In this case, we would multiply the number of combinations for choosing 5 from 7 sopranos (21) by the number of combinations for choosing 4 from 9 altos (126.) This gives us 21 * 126 = 2646 different combinations.