Permutations and Combinations: chosing 3 consonants from BANANA SPLIT

Miss_Hickey

New member
Joined
Dec 28, 2017
Messages
5
I am a high school Maths teacher seeking help with permutations and combinations. My question is:

Three letters are chosen at random from the word BANANA SPLIT. What is the probability that they are all consonants?

The worked solution uses combinations as follows:

There are 11 letters and 7 consonants.

consonant combinations = 7C3
= 7!/{3!(7-4)!}
= 35

total combinations = 11C3
= 11!/{3!(11-3)!}

P(all consonants) = 35/165
= 7/33

I understand all of the steps, except for why combinations is chosen rather than permutations. It is my understanding that combinations are used when the order of choices does not matter. However, in probability the order would matter - for example, the choice of SPL from the letters should be counted along with PSL, PLS, SLP etc.

When I calculated the example using permutations, I did not get 7/33:


consonant permutations = 7P3​
= 7!/3!​
= 840​


total permutations = 11P3​
=11!/3!​
= 6 652 800​


P(all consonants) = 840 / 6 652 800​
= 1/7920

Can someone please explain my error in reasoning? I am very confused about why combinations is used instead of permutations.

Thank you!
 
You'll have to explain why you think order matters. Nothing in the problem statement suggests it.

It is only a coincidence that the letters spell "Banana Split". Once you throw them into a hat for the random draw, they don't spell anything.
 
In light of Dr. P's answer below, I am retracting my original post here as not helpful at best and confusing at worst.
 
Last edited:
I am a high school Maths teacher seeking help with permutations and combinations. My question is:

Three letters are chosen at random from the word BANANA SPLIT. What is the probability that they are all consonants?

The worked solution uses combinations as follows:

There are 11 letters and 7 consonants.

consonant combinations = 7C3
= 7!/{3!(7-4)!}
= 35

total combinations = 11C3
= 11!/{3!(11-3)!}

P(all consonants) = 35/165
= 7/33

I understand all of the steps, except for why combinations is chosen rather than permutations. It is my understanding that combinations are used when the order of choices does not matter. However, in probability the order would matter - for example, the choice of SPL from the letters should be counted along with PSL, PLS, SLP etc.

When I calculated the example using permutations, I did not get 7/33:

consonant permutations = 7P3
= 7!/3!​
= 840​

total permutations = 11P3​
=11!/3!​
= 6 652 800​

P(all consonants) = 840 / 6 652 800​
= 1/7920

Can someone please explain my error in reasoning? I am very confused about why combinations is used instead of permutations.

Thank you!

Both permutations and combinations can be used for a problem like this, because you can count either the sets you could choose, or the sequences you could choose; and I think it is reasonably clear that you are taking the "letters" as if they were distinct tiles with those letters written on them (not worrying about duplicates), so the basic idea of what you did is correct.

All you did wrong was to calculate the permutations incorrectly. It should be:

consonant permutations = 7P3
= 7!/(7-3)! = 7*6*5
= 210

total permutations = 11P3
=11!/(11-3)! = 11*10*9
= 990

P(all consonants) = 210 / 990
= 7/33​

Note that the ratio of combinations is

\(\displaystyle \dfrac{C(7,3)}{C(11,3)} = \dfrac{7!/[(7-3)!3!]}{11!/[(11-3)!3!]} = \dfrac{7!/(7-3)!}{11!/(11-3)!} = \dfrac{P(7,3)}{P(11,3)}\)

which shows why both results must be the same.
 
Extension Question

Thank you so much for your help.

Is it also possible to do this question using both permutations and combinations? The texbook did it with combinations, but I do not understand why.

Four marbles are taken at random from a bag containing 5 red, 3 green and 4 yellow marbles. What is the probability that two red and two yellow marbles are taken?


It was my understanding that all possible arrangements of red marbles needed to be accounted for. Permutations does this. However, combinations doesn't account for the order in which the marbles are drawn.

If the first red marble is R_1 and the second is R_2, permutations would count both R_1, R_2 and R_2, R_1. However, combinations would only count this once. Isn't it important to consider the order in probability, as both are a possible outcome?

By combinations:

Combinations of two red marbles = 5C2​
= 5!/[2!(5-2)!]​
= 10​

Combinations of two yellow marbles = 4C2​
= 4!/[2!(4-2)!]​
= 6​

Combinations of four marbles = 12C4​
= 12!/[4!(12-4)!]​
= 495​

Combinations of two red and two yellow marbles = 5C2 x 4C2​
= 10 x 6​
= 60​

P(2 red and 2 yellow) = 60/495​
= 4/33​

By permutations

Permutations of two red marbles = 5P2​
= 5!/[(5-2)!]​
= 20​

Permutations of two yellow marbles = 4P2​
= 4!/[(4-2)!]​
= 12​

Permutations of four marbles = 12P4​
= 12!/[(12-4)!]​
= 11 880​

Permutations of two red and two yellow marbles = 5P2 x 4P2​
= 20 x 12​
= 240​

P(2 red and 2 yellow) = 240/11880​
= 2/99​
 
Because the problem itself does not refer to order, either permutations or combinations may be used. But you have to be consistent.

If you do the work with combinations, everything comes easily, because the question is really about combinations. (The combination formula makes all necessary adjustments for different orderings.)

With permutations, you have to think about permutations throughout. The number of ways to ORDER two red and two yellow is NOT the product of the numbers of ways to order two red, and the number of ways to order two yellow. That would only produce permutations of the form RRYY. You have to also consider different ways to order the two colors mixed together. One way to do this is to consider the number of ways to choose two positions in which to put the reds, 4C2 = 6. The answer is therefore

P(2 red and 2 yellow) = [5P2 * 4P2 * 4C2] / 12P4 = [20 * 12 * 6] / 11,880 = 1440/11,880 = 4/33

So you do get the same answer either way.
 
It was my understanding that all possible arrangements of red marbles needed to be accounted for. Permutations does this. However, combinations doesn't account for the order in which the marbles are drawn.
In an earlier answer, Dr. P said you can think about sets or you can think about sequences so long as you are consistent. Using the word "arrangements" does not clarify that distinction.

You have twelve marbles. Each marble has two numbers on it, a white number and a black number. The white numbers run from 1 through 12. Five marbles are red, numbered 1 through 5 in black. Four marbles are yellow, numbered 1 through 4 in black. And three marbles are green, numbered 1 through 3 in black.

How many different sequences of four white numbers can I form?
Obviously, 12 * 11 * 10 * 9.

How many different sequences of two black numbers can I form from the red marbles?
Obviously, 5 * 4.

How many different sequences of two black numbers can I form from the yellow marbles?
Obviously, 4 * 3.

how about sequences of two red and two yellow. If I start with a red, the second red can come in any of 3 positions. If I start with a yellow, the second yellow can come in any of 3 positions. So there are six distinct patterns of red and yellow

rryy, ryry, ryyr, yyrr, yryr, and yrry.

So my number of sequences satisfying the conditions is 6 * 5 * 4 * 4 * 3.

The probability is:

\(\displaystyle \dfrac{6 * 5 * 4 * 4 * 3}{12 * 11 * 10 * 9} = \dfrac{2 * 3 * 5 * 4}{11 * 10 * 9} = \dfrac{4}{33}.\)

But the question was not interested in sequences. It was interested in sets, in collections.

How many distinct collections of 2 red marbles can you form from 5?

\(\displaystyle \dfrac{5!}{(2!)(3!)} = \dfrac{5 * 4}{2} = 10.\)

How many distinct collections of 2 yellow marbles can you form from 4?

\(\displaystyle \dfrac{4!}{(2!)(2!)} = \dfrac{4 * 3 }{2} = 6.\)

How many distinct collections of four marbles can you form from 12?

\(\displaystyle \dfrac{12!}{(4!)(8!)} = \dfrac{12 * 11 * 10 * 9}{4 * 3 * 2} = 11 * 5 * 9 = 495.\)

Probability is:

\(\displaystyle \dfrac{6 * 10}{495} = \dfrac{60}{495} = \dfrac{4 * 15}{33 * 15} = \dfrac{4}{33}.\)

Short answer: don't think about sequences unless you must, and if you must, count every sequence.

It is easy to imagine distinguishing by what came first, what came second. That makes thinking sequentially seem intuitive. It is in fact very tricky because it is easy to forget a relevant sequence.
 
Last edited:
Top