Using the combination formula, it is found that:
1. A. 7 combinations are possible.
B. 21 combinations are possible.
C. 1 combination is possible.
2. There are 245 ways to group them.
<h3>What is the combination formula?</h3>
is the number of different combinations of x objects from a set of n elements, given by:
![C_{n,x} = \frac{n!}{x!(n-x)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
Exercise 1, item a:
One letter from a set of 7, hence:
![C_{7,1} = \frac{7!}{1!6!} = 7](https://tex.z-dn.net/?f=C_%7B7%2C1%7D%20%3D%20%5Cfrac%7B7%21%7D%7B1%216%21%7D%20%3D%207)
7 combinations are possible.
Item b:
Two letters from a set of 7, hence:
![C_{7,2} = \frac{7!}{2!5!} = 21](https://tex.z-dn.net/?f=C_%7B7%2C2%7D%20%3D%20%5Cfrac%7B7%21%7D%7B2%215%21%7D%20%3D%2021)
21 combinations are possible.
Item c:
7 letters from a set of 7, hence:
![C_{7,7} = \frac{7!}{0!7!} = 1](https://tex.z-dn.net/?f=C_%7B7%2C7%7D%20%3D%20%5Cfrac%7B7%21%7D%7B0%217%21%7D%20%3D%201)
1 combination is possible.
Question 2:
Three singers are taken from a set of 7, and four dances from a set of 10, hence:
![T = C_{7,3}C_{10,4} = \frac{7!}{3!4!} \times \frac{10!}{4!6!} = 245](https://tex.z-dn.net/?f=T%20%3D%20C_%7B7%2C3%7DC_%7B10%2C4%7D%20%3D%20%5Cfrac%7B7%21%7D%7B3%214%21%7D%20%5Ctimes%20%5Cfrac%7B10%21%7D%7B4%216%21%7D%20%3D%20245)
There are 245 ways to group them.
More can be learned about the combination formula at brainly.com/question/25821700