The coach of a junior league baseball team must choose 3 players to play as fielders from 9 young people who have tried out. In how many ways can these three players be chosen?
1400/121 = ~11.5702...
Which is over 11. So yes, you have enough.
A baseball team has 1,400 to buy baseball uniforms. Each uniform costs $124. Does the team have enough money to buy 11 baseball uniform.
Well, choosing a junior league baseball team is no joke! But let me clown around with the numbers to find the answer for you.
To choose 3 players from 9, we can use a combination formula. The formula goes like this:
nCr = n! / r! * (n - r)!
Where n is the total number of players (9 in this case) and r is the number of players we want to choose (3 in this case).
Plugging the numbers into the formula, we get:
9C3 = 9! / 3! * (9 - 3)!
= 9! / 3! * 6!
= (9 * 8 * 7 * 6!) / (3 * 2 * 1 * 6!)
= (9 * 8 * 7) / (3 * 2 * 1)
= 84
Voila! There are 84 ways to choose 3 players from a group of 9. Now, that's a lot of potential clown teammates on the field!
To find the number of ways to choose 3 players from a group of 9, we can use the concept of combinations. The number of combinations is calculated using the formula:
C(n, r) = n! / (r!(n - r)!)
Where n represents the total number of players and r represents the number of players to be chosen (in this case, 3). The exclamation mark (!) denotes factorial, which means multiplying a number by all positive integers less than it down to 1.
In this question, we have 9 players to choose from and we want to choose 3 players. Plugging these values into the combination formula, we get:
C(9, 3) = 9! / (3!(9 - 3)!)
Simplifying:
C(9, 3) = 9! / (3! * 6!)
Next, we evaluate the factorials:
9! = 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1
3! = 3 * 2 * 1
6! = 6 * 5 * 4 * 3 * 2 * 1
Substituting these values back into the expression:
C(9, 3) = (9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1) / ((3 * 2 * 1) * (6 * 5 * 4 * 3 * 2 * 1))
The common factors in the numerator and denominator cancel out:
C(9, 3) = (9 * 8 * 7) / (3 * 2 * 1)
Now, we calculate the resulting expression:
C(9, 3) = 504 / 6
C(9, 3) = 84
Therefore, there are 84 ways in which the coach can choose 3 players to play as fielders from a group of 9.
combinations of 9 taken 3 at a time
= 9!/[ (9-3)!(3!)]
= 9*8*7/(3*2)
= 3*4*7 = 12*7 = 84