Imagine that you have ten small light bulbs, of which three are burned out a) in how many ways can you randomly select 4 of the light bulb b) in how many ways can you select two good light bulbs and two burned out light bulbs?
Hel needed Urgently
a) In order to find the ways you can randomly select 4 of them, we use Combinations:
C(10,4)
= 10!/6!4! = 210
b) Two select in a specific way, we use different combinations for the burnt out and good bulbs, and multiply them:
C(7,2) x C(3,2)
= 7!/5!2! x 3!/2!
= 7 x 3 x 3
= 63
To solve these problems, we will use the concept of combinations.
a) In how many ways can you randomly select 4 light bulbs?
To find the number of ways to select 4 light bulbs out of a set of 10, we can use the combination formula. In this case, 3 bulbs are burned out, meaning there are only 7 good bulbs to choose from.
The formula for combinations is:
C(n, r) = n! / (r! * (n - r)!)
Where n is the total number of items and r is the number of items to be chosen.
Using the formula, we have:
C(7, 4) = 7! / (4! * (7 - 4)!)
= 7! / (4! * 3!)
= (7 * 6 * 5 * 4!) / (4! * 3 * 2 * 1)
= (7 * 6 * 5) / (3 * 2 * 1)
= 35
Therefore, there are 35 ways to randomly select 4 light bulbs from a set of 10 when 3 are burned out.
b) In how many ways can you select 2 good light bulbs and 2 burned out light bulbs?
To solve this problem, we need to choose 2 good bulbs from the remaining 7 good bulbs and 2 burned out bulbs from the set of 3.
The formula for calculating combinations would be:
C(7, 2) * C(3, 2)
Using the combination formula, we have:
C(7, 2) = 7! / (2! * (7 - 2)!)
= (7 * 6) / (2 * 1)
= 21
C(3, 2) = 3! / (2! * (3 - 2)!)
= 3! / (2 * 1)
= 3
Multiplying the combinations together, we get:
C(7, 2) * C(3, 2) = 21 * 3 = 63
Therefore, there are 63 ways to select 2 good light bulbs and 2 burned out light bulbs.