Mac J.

asked • 05/21/15

How many combinations is their to get the word banana. You can use a letter many times when doing so

 B A N
 A N A
 

1 Expert Answer

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Stephanie M. answered • 05/22/15

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Start by figuring out how many total ways there are to rearrange the letters B, A, N, A, N, and A.

 

There are 6 possible letters to choose for the first position, 5 remaining for the second spot, 4 for the third spot, 3 for the fourth spot, 2 for the fifth spot, and 1 for the sixth spot. That's a total of:

 

6! = 6×5×4×3×2×1 = 720 ways

 

But we've double-counted a few combinations... BANANA (where the first A is A1) and BANANA (where the first A is A2) are identical. We'll need to divide by the number of ways there are to rearrange just the A's and the number of ways there are to rearrange just the N's, since we want to treat all the A's the same and all the N's the same.

 

There are 3×2×1 = 6 ways to rearrange the A's and 2×1 = 2 ways to rearrange the N's. So, there are:

 

720/(6×2) = 720/12 = 60 ways to rearrange all the letters in BANANA.

 

 


Next, figure out how many of those arrangements are the word "BANANA." Since we've already taken duplicate letters (and therefore duplicate arrangements) into account, there's actually only 1 arrangement that spells the word BANANA.




So, in total, there are 60 ways to rearrange the letters in BANANA, only 1 of which is the word BANANA. That gives us a 1/60 probability of actually getting the word BANANA.

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Dan R.

Since Google sent me here, I feel the urge to correct this so others don't fall into the same mistake. The probability of arranging BANANA in the correct spelling order is 1/60. The mistake that she made is that in the numerator she is treating repeated letters as distinguishable, and in the denominator she is treating repeated letters as indistinguishable (and thus dividing by 12 to compensate). You need to stick one method, since there is one sample space for the probability of an event, i.e. a distinguishable or indistinguishable sample space. Interestingly enough, it occurs to me that the probability comes out the same whether you count repeated letters as distinguishable, 12/720, or indistinguishable, 1/60. To make it clearer, solve the simpler problem of finding the probability of getting the BAA arrangement, given the three letters {A,A,B}, using the distinguishable sample space { aAB , AaB, ABa, aBA, BAa, BaA } versus the indistinguishable one { AAB, ABA, BAA}.
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03/25/24

Stephanie M.

tutor
Hi Dan, Thank you for catching my mistake! I wrote this answer nearly a decade ago, and I clearly wasn't careful enough with my math when I did. You're absolutely right: I should've either taken the duplicate letters into account in the numerator, or ignored them in the denominator. I'll correct my original answer, but wanted to leave this response here so your comment won't seem out of place. One thing I could have done to catch my OWN mistake is perform a simple logic check once I'd gotten my answer: Does 1/5 seem like a reasonable probability for getting the word BANANA? The answer is clearly "no." The probability of just the B ending up in the right place is 1/6, already lower than 1/5 before even taking the A's and N's into account. Just thinking through that would have indicated that I must've done something wrong along the way, and would have given me the opportunity to work through the problem more carefully. Thanks again for the correction! Hopefully the updated answer will save future Googlers from making the same mistake. =)
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03/25/24

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