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Using the letters in the word ADDITION, find the number of permutations that can be formed using 3 letters at a time.
rikhilg:
84 is the answer
I dont think so fam
choices are A.300B.
323
C.
336
D.
345
323
C.
336
D.
345
A:300 B:323 C:336 or D:345
Answers
Answered by
1
Answer:
84
Step-by-step explanation:
Hi Buddy! Nice to see this question and help you!!
Here's the answer:
Well, first treat each D and I to be different.
The the permutations will be ⁸P₃ right?
But there are 2 D and 2 I. Use the formula: P/ n! m!
Where n and m are the number of times repeated letters are appearing, and P is the permutations ignoring the repititions.
I appears 2 times. D appears 2 times. Hence n=2 and m=2
So your answer is : ⁸P₃/2*2
=8!/3!*2*2
=8x7x6/2x2
=8x7x6/4
=2x7x6
=14x6
=84.
There you go!!
Hope it helps, and pls mark brianliest. Thanks Buddy! Bye!
dude how? no way! I'm sure of my asnwer
go ahead and check the net
Been through there and tried it on my own, i think the question is messed up lol
gonna guess and pray
ya... Best of luck buddy!!!
If anyone else has this problem, i guessed and got it right, it was C: 336
can't be 336 man!!! has to be 84
notice that 84 x 4 = 336.
The question is wrong! lol!
Whoever made the question forgot that there are repititions of letters. lol!!
Answered by
0
HEY MATE HERE IS YOUR ANSWER
The number of permutations if all letters were distinct = 6P36P3.
As 'S' and 'E' are repeating, the arrangement in which 'S' are interchanged are the same, so we divide by 2!2! for 'S' and likewise for 'E'.
Number of permutations =6P32!.2!=30=6P32!.2!=30
Which isn't the correct answer. I would like to get some help here.
The number of permutations if all letters were distinct = 6P36P3.
As 'S' and 'E' are repeating, the arrangement in which 'S' are interchanged are the same, so we divide by 2!2! for 'S' and likewise for 'E'.
Number of permutations =6P32!.2!=30=6P32!.2!=30
Which isn't the correct answer. I would like to get some help here.
The number of permutations if all letters were distinct = 6P36P3.
As 'S' and 'E' are repeating, the arrangement in which 'S' are interchanged are the same, so we divide by 2!2! for 'S' and likewise for 'E'.
Number of permutations =6P32!.2!=30=6P32!.2!=30
Which isn't the correct answer. I would like to get some help here.
The number of permutations if all letters were distinct = 6P36P3.
As 'S' and 'E' are repeating, the arrangement in which 'S' are interchanged are the same, so we divide by 2!2! for 'S' and likewise for 'E'.
Number of permutations =6P32!.2!=30=6P32!.2!=30
Which isn't the correct answer. I would like to get some help here.
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