<ol>
<li>The acme plumbing company will send a team of 3 plumbers to work on a certain job. The company has 4 experienced plumbers and 4 trainees. If a team consists of 1 experienced plumber and 2 trainees, how many different such teams are possible?</li>
</ol>
<p>I keep getting 48... but the answer is 24... wth</p>
<p>4 x 4C2 because the order of the trainees don’t matter</p>
<p>Method #1:
Let numbers = Experienced Plumbers
Let letters = Trainee Employees</p>
<p>1AB
1AC
1AD
1BC
1BD
1CD</p>
<p>2AB
2AC
2AD
2BC
2BD
2CD</p>
<p>3AB
3AC
3AD
3BC
3BD
3CD</p>
<p>4AB
4AC
4AD
4BC
4BD
4CD</p>
<p>Method #2:
ABCD have a combination of 6 or 3! (3 x 2 x 1).
If it was ABCDE would have a combination of 24 or 4! (4 x 3 x 2 x 1).
6 x 4 (number of trained employees) = 24</p>
<p>You are doing a permutation. i.e. 4 plumbers<em>4 trainees</em>3 trainees.</p>
<p>However, you should do a combination since Jeff with trainees Stan and Casey is the same as Jeff with trainees Casey and Stan. So divide by two.</p>
<p>do you always divide by to when order doesn’t matter? so i can solve via permutations then simply divide by 2?</p>
<p>When order doesn’t matter, you divide by the number of orders (in this case 2, but not always 2). </p>
<p>The number of orders for n slots is of course n!. So you divide by n!.</p>
<p>Orders as in you only choose 2 trainees? so if i chose 3 trainees? it would be divide by 3?</p>
<p>Order doesn’t matter as in, for the three trainees, no “President, Vice President, Secretary” which are 3 different positions under which there would be more combination. All three are of the same level, so in that case you use C instead of P when doing the calculations. So think of it as 4C1 x 4C2- just use your scientific calculator for this.</p>
<p>why is it 4 C1 x 4C2? if order doesn’t matter for example like 3 trainnees? or 2 trainees?</p>
<p>bump10char</p>
<p>When solving this problem think about it in two parts. The first part is what is the combination of 4 the trainee plumbers. The second part is multiplying that combination by 4 since each plumber is paired with the combination.</p>
<p>Combination: (6 total) -> if order doesn’t matter = combination
</p>
<p>Permutation: (12 total) -> if order matters = permutation
</p>
<p>Mathematically using Factorials:
= (n!)(k!*(n-k)!)</p>
<p>n = total number of trainee plumbers
k = amount used or number of trainees needed (if 3 trainee plumbers were paired with an experienced plumber it would be 3 instead of 2)</p>
<p>= (4!)/(2!*2!)
= (4 * 3 * 2 * 1)/(2 * 1 * 2 * 1)
= 24/4
= 6</p>
<p>= (combination) * (how many times the combination will be used)
= (6) * (4)
= 24</p>
<p>i see… now if i wanted to use calculator functions how would i do it?</p>
<p>Math->PRB->nCr or Math->PRB->!</p>
<p>so for this problem it would be 4 C 2? then multiply by 4 ? if i chose 3 trainees out of 4, it would be 4 C 3?</p>
<p>Yes, the nCr function on your calculator simply does: (n!)/(k!*(n-k)!)</p>
<p>4nCr2 = 6
(4!)/(2!*(4-2)!) = 6</p>
<p>4nCr3 = 4
(4!)/(3!*(4-3)!) = 4</p>
<p>ahhh so this works with any combination problem ? can u give me an example one that i can do? lol</p>