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2. How many 3. In a 10-item Mathematics problem-solving test, how many ways can you select 5 problems to solve? Grom a class of 30 students of orade 7 in how many ways can a teacher makes a group with A​

Sagot :

Answer:

252.

Step-by-step explanation:

Order is not important because answering problem 1,2,3,4 and 5 is the same as answering problems 4,3,5,2 and 1. Since the order is not important, we use combination. The formula for taking r things from n possible ones with no order is expressed by.

\frac{n!}{n!(n-r)!}

n!(n−r)!

n!

We have 5 items to choose from 10 possible problems, so r = 5, n = 10. Substituting it to the formula and simplifying gives us:

\begin{gathered}=\frac{10!}{5!(10-5)!}\\\\=\frac{10!}{5!5!}\\ \\=\frac{10*9*8*7*6*5!}{5*4*3*2*1*5!}\\ \\=3*2*7*6\\\\=252\end{gathered}

=

5!(10−5)!

10!

=

5!5!

10!

=

5∗4∗3∗2∗1∗5!

10∗9∗8∗7∗6∗5!

=3∗2∗7∗6

=252

There are 252 ways to select 5 problems from 10 possible problems.