0
What else can I help you with?
How many ways can I select a committee of 3 people from 10 people?
To select a committee of 3 people from 10, you can use the combination formula ( C(n, k) = \frac{n!}{k!(n-k)!} ). Here, ( n = 10 ) and ( k = 3 ). This gives ( C(10, 3) = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 ). Therefore, there are 120 ways to select a committee of 3 people from 10.
How many ways can you select 3 marbles from a jar containing 10 different colored marbles?
To select 3 marbles from a jar containing 10 different colored marbles, you can use the combination formula ( C(n, r) = \frac{n!}{r!(n-r)!} ). Here, ( n = 10 ) and ( r = 3 ). Thus, the number of ways to select the marbles is ( C(10, 3) = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 ). Therefore, there are 120 different ways to select 3 marbles.
How many ways can a student select 7 questions from an exam containing 11 questions if he must answer the last question?
If he must answer the last question, he effectively needs to select 6 from 10. This can be done in 10C6 = 10*9*8*7/(4*3*2*1) = 210 ways.
How many ways can a student select 5 questions from an exam containing twelve questions if he must select the last question?
If he must include the last question, then his choices really boil down tohow many ways he can select 4 more from the remaining 11.The number of ways he can select 4 from 11 is (11 x 10 x 9 x 8) = 7,920 ways .But each group of 4 can be selected in 24 different ways.So he can only wind up with 7920/24 = 330 different groups of questions.
How many ways are there to assign the 10 positions by selecting players from the 13 people who show up?
To assign 10 positions from 13 players, you can use the concept of permutations since the order of selection matters. The number of ways to select and arrange 10 players out of 13 is given by the permutation formula ( P(n, r) = \frac{n!}{(n-r)!} ). For this scenario, it would be ( P(13, 10) = \frac{13!}{(13-10)!} = \frac{13!}{3!} ), resulting in 13,860 different ways to assign the positions.
How many ways can I select a committee of 3 people from 10 people?
To select a committee of 3 people from 10, you can use the combination formula ( C(n, k) = \frac{n!}{k!(n-k)!} ). Here, ( n = 10 ) and ( k = 3 ). This gives ( C(10, 3) = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 ). Therefore, there are 120 ways to select a committee of 3 people from 10.
How many ways can Aquiel select 4 books to buy form the 10 best sellers and read them?
Aquiel can select 4 books out of 10 best sellers, which can be done in 10 choose 4 ways, also denoted as "10C4" or "10 choose 4". This would result in 210 different ways for Aquiel to select and read the 4 books.
A student has an exam with 30 questions. They are to select 10 of these questions to answer. In how many ways can they complete this exam?
they can complete this exam 300 ways
How many ways can you select 3 marbles from a jar containing 10 different colored marbles?
To select 3 marbles from a jar containing 10 different colored marbles, you can use the combination formula ( C(n, r) = \frac{n!}{r!(n-r)!} ). Here, ( n = 10 ) and ( r = 3 ). Thus, the number of ways to select the marbles is ( C(10, 3) = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 ). Therefore, there are 120 different ways to select 3 marbles.
How many ways can a student select 7 questions from an exam containing 11 questions if he must answer the last question?
If he must answer the last question, he effectively needs to select 6 from 10. This can be done in 10C6 = 10*9*8*7/(4*3*2*1) = 210 ways.
In how many ways can a committee of three men and four women be formed from a group of 10 men and 10 women?
I'm going with 25,200 3 men out of 10 may be chosen in 10C3 ways = 10 ! / 3! 7 ! = 120 ways. 4 women may be chosen out of 10 in 10C4 = 10 ! / 4! 6! ways = 210 ways. Therefore, a committee with 3 men and 4 women can be formed in 120 x 210 = 25,200 ways.
How many ways can a student select 5 questions from an exam containing twelve questions if he must select the last question?
If he must include the last question, then his choices really boil down tohow many ways he can select 4 more from the remaining 11.The number of ways he can select 4 from 11 is (11 x 10 x 9 x 8) = 7,920 ways .But each group of 4 can be selected in 24 different ways.So he can only wind up with 7920/24 = 330 different groups of questions.
How many ways can a dinner patron select 3 appetizers and 2 vegetables if there are 6 appetizers and 5 vegetables on the menu?
Oh, what a delightful question! To find the total ways a dinner patron can select 3 appetizers out of 6, we use combinations, which is like a recipe for choosing items without regard to the order. So, the dinner patron can choose 3 appetizers out of 6 in 20 ways. Similarly, they can choose 2 vegetables out of 5 in 10 ways. So, the total ways to select 3 appetizers and 2 vegetables is 20 * 10 = 200 ways. Isn't that just lovely?
How many ways can you make 1 dollar using 5 cents 20 cents and 50 cents?
10 ways.10 ways.10 ways.10 ways.
What is the probability of 1 to 50 at random selecting numers with 4 in the 10 place?
The answer depends on how many numbers you select!The answer depends on how many numbers you select!The answer depends on how many numbers you select!The answer depends on how many numbers you select!
How many ways are there to assign the 10 positions by selecting players from the 13 people who show up?
To assign 10 positions from 13 players, you can use the concept of permutations since the order of selection matters. The number of ways to select and arrange 10 players out of 13 is given by the permutation formula ( P(n, r) = \frac{n!}{(n-r)!} ). For this scenario, it would be ( P(13, 10) = \frac{13!}{(13-10)!} = \frac{13!}{3!} ), resulting in 13,860 different ways to assign the positions.
How many ways can 5 people be selected out of 10?
The number of ways is 10C5 = 10!/(5!*5!) = 10*9*8*7*6/(5*4*3*2*1) = 252
