First, we will see how many things we have to select from the total. After that, we will solve the question using the above formula.
Complete step by step answer:
Let us solve this question.
This is a question of permutation and combinations.
Let us first know what permutations and combinations.
A number of permutations [order matters] of n things taken r at a time:
\[{}^{n}{{P}_{r}}=\dfrac{n!}{\left[ n-r \right]!}\]
A number of combinations [order does not matter] of n things taken r at a time:
\[{}^{n}{{C}_{r}}=\dfrac{n!}{\left[ n-r \right]!r!}\]
So, in this question, we have to find that in how many ways, we can select a committee of five members from a group of 10 people.
So, in this question, the process of choosing the persons does not matter which means the order of choice does not matter.
So, we are going to use the formula for this question is \[{}^{n}{{C}_{r}}=\dfrac{n!}{\left[ n-r \right]!r!}\]
We are choosing 5 persons from 10 persons where order does not matter.
Then, the numbers of ways are \[\dfrac{10!}{[10-5]!5!}\]
As we know that \[n!=n\times [n-1]\times [n-2]\times [n-3]\times ........\times 2\times 1\]
Then, \[\dfrac{10!}{[10-5]!5!}\] can be written as
\[\dfrac{10!}{[10-5]!5!}=\dfrac{10!}{5!\times 5!}=\dfrac{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{\left[ 5\times 4\times 3\times 2\times 1 \right]\times \left[ 5\times 4\times 3\times 2\times 1 \right]}\]
Which is also can be written as
\[\dfrac{10!}{[10-5]!5!}=\dfrac{10\times 9\times 8\times 7\times 6}{5\times 4\times 3\times 2\times 1}=\dfrac{10}{5}\times 9\times \dfrac{8}{4}\times 7\times \dfrac{6}{3\times 2}=2\times 9\times 2\times 7\times 1=252\]
Hence, \[\dfrac{10!}{[10-5]!5!}=252\]
Therefore, the number of ways of selecting a committee of 5 members from a group of 10 persons is 252.
Note:
For solving this type of question, we should have a better knowledge of permutation and combinations. And, also remember the formulas of permutation and combination. Make sure that no calculation mistakes have been done in solving the question. Otherwise, the solution will be wrong. For example, \[\dfrac{10!}{[10-5]!5!}=\dfrac{10\times 9\times 8\times 7\times 6}{5\times 4\times 3\times 2\times 1}=\dfrac{10}{5}\times 9\times \dfrac{8}{4}\times 7\times \dfrac{6}{3\times 2}=2\times 9\times 2\times 7\times 1=252\]
In the above, mistakes can be made during calculations.
Algebra Systems of Equations and Inequalities Probability and Combinations
1 Answer
hev1
Mar 24, 2018
There are#252#ways to select a committee of five members from a group of#10#people.
Explanation:
The number of ways to select#b#people from a total of#a#people is#[a!]/[b!*[a-b]!#.
Plugging in#10#for#a#and#5#for#b#:
#P = [10!]/[5![10-5]!]#
#P = 3628800/[120*5!]#
#P = 3628800/14400#
#P = 252#
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Question 9147: I need help on finding the number of different starting teams if five basketball players are chosen from a team of ten players
Answer by DWL[56] [Show Source]:
You can put this solution on YOUR website!
This is a permutation and combination problem. I'll Give you the answer, but you need to read about these two subjects.
Permutation Part
Now we have 10 players and want to make groupings of 5 people. It's harder to list all those permutations. To find the number of five-people permutations that we can make from 10 people without repeated [10_P_5], we'd like to have a formula because there are 30,240 such permutations and we don't want to write them all out!
For five-people permutations, there are 10 possibilities for the first person, 9 for the second, 8 for the third, and 7 for the fourth, and 6 for the last person. We can find the total number of different five-people permutations by multiplying 10 x 9 x 8 x 7 x 6 = 30,240. This is part of a factorial
To arrive at 10 x 9 x 8 x 7 x 6, we need to divide 10 factorial [10 because there are ten objects] by [10-5] factorial [subtracting from the total number of objects from which we're choosing the number of objects in each permutation]. You can see below that we can divide the numerator by 5 x 4 x 3 x 2 x 1:
10_P_5 =
Combination Part
we have 10 people from which we wish to choose 5 and we want to find the number of combinations of size 5 without repeated people that can be made from the ten people. To calculate 10_C_5, which is 120, we don't want to have to write all the combinations out!
Since we already know that 10_P_5 = 30,240, we can use this information to find 10_C_5. Let's think about how we got that answer of 30240. We found all the possible combinations of 5 that can be taken from 10 [10_C_5]. Then we found all the ways that five people in those groups of size 5 can be arranged: 5 x 4 x 3 x 2 x 1 = 5! = 120. Thus the total number of permutations of size 5 taken from a set of size 10 is equal to 5! times the total number of combinations of size 5 taken from a set of size 10: 10_P_5 = 5! x 10_C_5.
When we divide both sides of this equation by 5! we see that the total number of combinations of size 5 taken from a set of size 10 is equal to the number of permutations of size 5 taken from a set of size 10 divided by 5!. This makes it possible to write a formula for finding 10_C_5:
10_C_5 =
So the answer is:
You can have 252 starting teams of five from a group of 10 players.