1. ## Correcting test

Questions that I need help correcting.

1) In how many ways can a six member committee form a subcommittee with at least two members?

2) A new theme park will have five roller coasters, six water rides, and eight suspension rides. How many different sets of rides could you try at this park?

3) What are the first three terms of the expansion of (x^2/7 - x/49)^6 ?

4) A project team of up to 5 people is being formed from a staff of 15. How many such teams are possible?

2. Have you tried any of these problems? It says "correcting test" so I assume these are problems you have seen already?

3. Yea I tried answering them but since its a test I don't have the correct answers for the ones I did wrong (my teacher doesn't correct them for us he just marks wrong), so I did more studying and tried them again and I just wanna check if I did it right.

1) 66 ways to make the sub committee of at least 2 members

2) 377 different sets of rides

3) I just couldn't get this one down as I couldn't understand how to apply the formula.

4) I got 1001

4. Hello, rozekruez!

Sorry, all wrong answers . . .

1) In how many ways can a 6-member committee form a subcommittee with at least 2 members?

"At least 2 members" means: 2, 3, 4, 5, or 6 members.

${6\choose2} + {6\choose3} + {6\choose4} + {6\choose 5} + {6\choose6}\;=\;15 + 20+15+6+1 \;=\;58$ ways.

2) A new theme park will have five roller coasters, six water rides, and eight suspension rides.
How many different sets of rides could you try at this park?
The problem is not clearly stated.

Assuming that we choose one of each type of ride,

. . there are: . $5\cdot6\cdot8 \:=\:240$ sets of rides.

3) What are the first three terms of the expansion of $\left(\frac{x^2}{7} - \frac{x}{49}\right)^6$ ?
We're expected to know the Binomial Expansion:

. . $(a + b)^n \;=\;a^n + na^{n-1}b + \frac{n(n-1)}{2}a^{n-2}b^2 + \hdots$

We have: . $a=\frac{x^2}{7},\;b=\frac{x}{7^2},\;n = 6$

Substitute: . $\left(\frac{x^2}{7}\right)^6 + 6\left(\frac{x^2}{7}\right)^5\left(\frac{x}{7^2}\r ight) + 15\left(\frac{x^2}{7}\right)^4\left(\frac{x}{7^2}\ right)^2$

. . . . . . . . $=\;\frac{x^{12}}{7^6} + 6\left(\frac{x^{10}}{7^5}\right)\left(\frac{x}{7^2 }\right) + 15\left(\frac{x^8} {7^4}\right)$ $\left(\frac{x^2}{7^4}\right)$

. . . . . . . . $= \;\frac{x^{12}}{7^6} + \frac{6x^{11}}{7^7} + \frac{15x^{10}}{7^8}$

4) A project team of up to 5 people is being formed from a staff of 15.
How many such teams are possible?

"Up to 5 people" means: 1, 2, 3, 4, or 5 people.

. . ${15\choose1} + {15\choose2} + {15\choose3} + {15\choose4} + {15\choose5} \;=$ . $15 + 105 + 455 + 1365 + 3003 \;=\;4943$