Combinations
How many arrangements are there of the letters of BANANA such that no two N's appear in adjacent positions?
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4 solutions
yeah, it's correct!
https://socratic.org/questions/how-do-you-calculate-permutations-of-a-word
how did you get 5!
This is probably the best answer but it could use more explanation. The word "BANANA" has 6-positions and uses 3 different letters with "A" repeating 3 times and N repeating twice. So, step 1 is calculating the number of arrangements of "BANANA" taking into account repeats by removing them. If there were no repeats, we would simply use 6!. To make this clear, we do not want to count: "BAAAN", "BAAAN" and "BAAAN" as 3 different words where each "A" is treated as if it were a different letter. We want to just count "BAAAN" once. To do this, we remove the permutations of repeats so we need to divide them out. 6!/(3!*2!). The denominator contains the permutations of repeats of "A" and multiplies (due to rule of product) by permutations of repeats of "N". That formula comes out to 60.
Part 2 of this problem is that we are introducing a restriction whereby the two "N"s cannot appear in adjacent positions. So, we want to calculate all the permutations where NN are adjacent and subtract them from our non-restricted formula with the arrangement of BANANA which is 60. To calculate the restriction, it's best to think of the NN as a single letter. So "B A A A NN" would be the equivalent of 5 letters (NN counts as 1). So now we need to find the permutations of this arrangement which is 5!/3!. The denominator of 3! takes into account the three "A" repeats. This comes out to 20, which we subtract from 60.
Final answer, therefore, is 60 - 20 = 40.
Wshss. How did you get this 5!/3!? i cannt understand tahat.
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See, again we have three A's repeating, thus it will be 5!/3!. Hope you get it
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sorry can you elaborate
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@Chaitnya Shrivastava – See, in the second step we are considering NN to be a single letter because we want to find out how many arrangements are there with both N adjesent to each other. So the total no of letters becomes 5 and the total no of letters repeating becomes 3 with that 3 A in BANANA. So 5!/3! We don't divide 5! By 2! Because we are considering the 2 N as one letter... I hope this clears your confusion :)
Thanks.... :D
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@Satvik Golechha – Why d'ya have to thank me for that? @Satvik Golechha
First count all arrangements with the letters of B A N A N A that are ( 3 6 ) ( 2 3 ) ( 1 1 ) = 6 0 . The we eliminate N N are ( 1 5 ) ( 3 4 ) = 2 0 words with N N chatacter. Thus answer is 6 0 − 2 0 = 4 0
(5 1) (4 3) is not 20 right
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5!/1!(5-1)! * 4!/3!(4-3)! = 5!/4! * 4!/3! = 5 * 4 = 20
Taking out the 2 N's we have "BAAA" which can be permuted in 4 ways. We can add the first N in 5 different ways and the second in 4. Divide by 2 because of repetition of N and you get
4 ∗ 5 ∗ 4 / 2 = 4 0
Arrangements of "BANANA" = 6!/(3! 2!) = 60
If we consider "NN" a single character, then we have 5!/3! = 20
60 - 20 = 40.