<p>How many different committees of 3 people each can Jim select from a group of 5 people?</p>
<p>Answer is 10 but I don't get how</p>
<p>Let’s assume the people are A, B , C , D , E
then you can pick:
ABC
ABD
ABE
ACD
ACE
ADE
BCD
BCE
BDE
CDE</p>
<p>And there you have 10 groups.
(note that BAC for example is the same as ABC, so you can’t count that twice!)
Just start each letter and continue the list untill you use the last 3 in a row.
Perhaps there’s a mathematic way to solve it, but that also works, but you must pay attention not to miss any, some people might get stuck after they finish with A.</p>
<p>This is the combination 5C3. You can do this in your graphing calculator in a few seconds. Type 5 followed ny nCr followed by 3 and then ENTER. The display will show 10.</p>
<p>In my graphing calculator, what is the difference between the !, nCr, and nPr functions?</p>
<p>n! is the same as nPn. For example, 5! = 5<em>4</em>3<em>2</em>1</p>
<p>P stands for permutation and C stands for combination.</p>
<p>5c3 = 5!/(3!.(5-3)!)
= 5.4/2.1
=5.2 =10</p>
<p>How do you know that that’s a combination? I thought order matters in this case so wouldn’t if be a permutation? And how do you know what goes to the left and right of the thing you put in the calculator</p>
<p>Order does not matter because ABC is the same committee as ACB and BCA, etc…</p>
<p>If instead of asking for a committee, they had asked for a president, vice-president and treasurer, that would make it a permutation question. ABC would no longer be the same as BCA. Order WOULD matter.</p>
<p>As for your calculator question: that’s just the standard format that your calculator expects: 10P3 is the number of arrangements of 3 items chosen from 10 possibilities. Bigger number goes first…</p>