3 questions: permutations and combinations

Question

Dear math experts,

I hope this finds you well.

I solved 2 questions and I wonder whether my answers are correct.

1) A coach must choose 8 players randomly from 10 players and assign them to three different cars to transport them to an out-of-town game: 2 in car A, 3 in car B, and 3 in car C. In how many ways can this be done?

In this case, we are choosing people randomly, therefore do we need to use combination and divide 10C8 by (2! 3! 3!), to avoid repetition?

2) In how many ways can the letters and numbers of MATH1011 be arranged, if all the letters must be put together?

In this case, we are putting each letter and number together, therefore I think that no repetition is allowed. Hence I think that: 4P4 x (4! / (3! 1!) is the method for this question.

If both methods are wrong, feel free to tell where are my errors and what are correct methods.

3) Lastly, there is a question that I saw in class but I did not understand the method for solving this problem.

Here is the question: An artist has created 20 original paintings, and she will exhibit some of them in 3 galleries. 4 paintings will be sent to gallery A, 4 to gallery B, and 3 to gallery C. In how many ways can this be done? 

For this question, I do not know why you include 9! in the denominator to solve this problem. Could you please elaborate? Also, why should the sum of the denominator be the same as numerator?

Thanks for your genuine assistance in advance and have a nice day.

Best regards,

A studious student willing to progress :)

Answer 1

To solve these problems, use the Fundamental Principle of Counting: Every arrangement can be done as series of steps. We count how many ways there are on each step, and then multiply these numbers.

1) A coach must choose 8 players randomly from 10 players and assign them to three different cars to transport them to an out-of-town game: 2 in car A, 3 in car B, and 3 in car C. In how many ways can this be done?

Step 1. Select 8 players from 10. There are 10C8 = 10!/(8!2!) ways.

Step 2. From the selected 8 players select 2 people in car A. There are 8C2 = 8!/(2!6!) ways

Step 3. From the rest 6 players select 3 people in car B. There are 6C3 = 6!/(3!3!) ways.

We do not need more steps because the rest 3 players should be assigned in car C by the only way.

Totally there are 10C8 * 8C2 * 6C3 = 10!/(8!2!) * 8!/(2!6!) * 6!/(3!3!) = 10! / (2!2!3!3!) ways.

2) In how many ways can the letters and numbers of MATH1011 be arranged, if all the letters must be put together?

Step 1. Select places to where we can put the letters. Because the letters must be put together, they can be put on the 1^st – 4^th places, 2^nd – 5^th places, …, 5^th – 8^th places. There are 5 ways to select the places.

Step 2. Arrange the 4 letters in the assigned places. There are 4! ways.

Step 3. Select to where put the digit “0”. 4 places are already selected for the letters, so there are 4 other places. There are 4 ways to put “0”.

We do not need more steps because the 3 digits “1” should be assigned to the 3 rest places by the only way.

Totally there are 5 * 4! * 4 = 480 ways.

3) An artist has created 20 original paintings, and she will exhibit some of them in 3 galleries. 4 paintings will be sent to gallery A, 4 to gallery B, and 3 to gallery C. In how many ways can this be done? 

Step 1. Select 4 paintings from 20 to gallery A. There are 20C4 = 20!/(4!16!) ways.

Step 2. Select 4 paintings from the rest 16 to gallery B. There are 16C4 = 16!/(4!12!) ways.

Step 3. Select 3 paintings from the rest 12 to gallery C. There are 12C3 = 12!/(3!9!) ways.

Totally there are 20C4 * 16C4 * 12C3 = 20!/(4!16!) * 16!/(4!12!) * 12!/(3!9!) = 20!/(3!4!4!9!) ways.