Math Destroyer Test 1 #38

[predental89]

How do you solve this :/ I am having the hardest time with permutations and combinations!! If someone can explain this I would appreciate it.

five chairs are lined up in a row. how many ways can 3 boys and 2 girls sit in these chairs if the girls always sit side by side?










Answer is D, 48 ways

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[jwnichols21]

Okay so, first of all this is a permutation problem because order MATTERS. When order matters, you just multiply the values out without dividing by anything.

Per chad's method:

Plot out the different situations.

Situation 1: G G B B B
Situation 2: B G G B B
Situation 3: B B G G B
Situation 4: B B B G G

Note that they said that the girls have to sit next to each other always. So these are the only 4 possible scenarios. Each scenario has equal probability, so I'm going to show you the probabilities for Situation 1 and then you just multiply that by 4.

Situation 1: 2 x 1 x 3 x 2 x 1

Explanation: Seat 1 has to be a girl. There are 2 possible girls. Seat 2 has to be a girl, but one has already sat in Seat 1, so only 1 possibility left. Seat 3 has to be a boy, 3 possibilities. Seat 4, boy, only 2 options left. And Seat 5, boy, 1 option left.

So when you multiply that out, you get 12.
Remember that there are 4 different possible Situations here, and embedded in each situation is the same numbers as above. So, 12 x 4 = 48.

Let me know if this helps!

 

[jwnichols21]

It works I think..but Chad advised not to use the equations b/c they're often a crutch and people don't know when to use which..I just listen to what Chad tells me 😀

 

[indigenoustw]

Why does the permutation equation that is in the front of Math Destroyer not work for this question?

The equation stated is P(n, r) = [n!/(n-r)!]

Can someone explain why this equation wouldn't work this type of problem?

Thanks

It works, but won't get you to the final answer.

Permutation only applies to arrangement of the same entities when the order matters.

In this case, since the girls have to sit together, they can be seen as one entity while the other 3 boys can be seen as another entity. So although there are 5 chairs, you can imagine that there are only 4 chairs for one group of 2 girls and 3 different boys to sit in. Remember, order matters in this case because chairs WILL be taken up by either the group of 2 girls or 3 boys. Therefore, how many ways can 3 boys and one group of girls sit in the 4 chairs?

One formula: "The number of permutations possible out of n objects (where p of one kind are alike, q of another kind are alike, ... and r of yet another kind are alike) can be expressed as:

n!/(p!q!...r!) where p + q + ... + r = n"


4!/(1!*3!), so 4 ways.

"4 ways" isn't the final answer because that only tells you how many ways can you rearrange the 2 different groups when there are 4 choices available. We still need to account for the number of ways each entity can be rearranged themselves.

For the girls, two seats for 2 girls, so 2P2 = 2 ways.
For the boys, three seats for 3 boys, so 3P3 = 6 ways.

Now, for each arrangement of the entities, there are 6 x 2 = 12 ways. Since there are 4 ways you can arrange the 2 entities, 4 x 12 = 48 ways, which is the final answer.

Hope it helped.

 

[predental89]

Okay so, first of all this is a permutation problem because order MATTERS. When order matters, you just multiply the values out without dividing by anything.

Per chad's method:

Plot out the different situations.

Situation 1: G G B B B
Situation 2: B G G B B
Situation 3: B B G G B
Situation 4: B B B G G

Note that they said that the girls have to sit next to each other always. So these are the only 4 possible scenarios. Each scenario has equal probability, so I'm going to show you the probabilities for Situation 1 and then you just multiply that by 4.

Situation 1: 2 x 1 x 3 x 2 x 1

Explanation: Seat 1 has to be a girl. There are 2 possible girls. Seat 2 has to be a girl, but one has already sat in Seat 1, so only 1 possibility left. Seat 3 has to be a boy, 3 possibilities. Seat 4, boy, only 2 options left. And Seat 5, boy, 1 option left.

So when you multiply that out, you get 12.
Remember that there are 4 different possible Situations here, and embedded in each situation is the same numbers as above. So, 12 x 4 = 48.

Let me know if this helps!

Thanks dude you explained it really well! 😎