Permutations with restrictions - items not together | ExamSolutions

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Permutations with restrictions.

In this tutorial I demonstrate how to calculate permutations (arrangements) where there are restrictions in place. In these examples certain items are not to be placed together. The examples used are:
1) In how many ways can 5 men and 3 women be arranged in a row if no two women are standing next to each other.
2) In how many ways can the letters in the word SUCCESS be arranged if no two S's are next to each other.

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vid is 13 years and this just cured my head migraine thanks doc

aeeeeeugh
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idk what i did whole year now i finally know the concept of permutation ... all teachers in class were useless but you sir, are a legend :D

marsalsarma
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God bless this man! explained way better than my teacher.

alexillo
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absolute legend! I have an awful stats teacher, but you have renewed my confidence in the subject. good stuff

BackyardRadio
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This was the hardest part of the CIE AS level Maths Statistics as it was very hard to comprehend why u could draw spaces between and apply the permutation and combination formula whilst believing that the n of n! or nCr is a fixed total number of a person or item

oneinabillion
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Thank you so much for the videos sir. They are incredibly helpful for me especially since that I am studying the topics by myself. I have a doubt and I will really appreciate if you could help me out with it.
For the first question, shouldn't the answer be multiplied by 3! because the 3 women can arrange themselves differently?

rinu
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Your method is correct.. sorry for doubting you I finally know why the other method doesn't work.

vincentzulu
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still out here saving lives 13 years later 🙏

jujtud
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Even in the second problem, we are obtaining a totally different answer by subtracting the number of ways where s' are together from the total number of ways:

SUCCESS:

Total arrangements possible= 7! / (3! . 2!)

Number of arrangements where s' are together = 5! / 2!
(Note: as no internal arrangement possible between s')

Number of ways where s' are not together = 420-60 =360.

ajmalrashid
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Can't thank you enough for this! made the chapter look so simple

huzaifaimran
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I know am the newest here .. the only thing that matter the 12 years old video helps ..old is gold great job

isaacnjuguna
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How do you figure out how many dashes to give before and after the numbers?

nafiskhandakar
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@ExamSolutions i hope u know that ur the man and a chad

gamingbraaa
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Hi Sir!

I tried to solve first principle by the method taught by you in the last
video. I am getting totally different answer. I rechecked it many a
times.

8 people (5 men+3 women). Total arrangements= 8! = 40320.
Number of arrangements where women are together= 6!*3! = 4320 (Note:
Considering women as a group, we have 6 groups and 3! internal
arrangement for 3 women). Number of arrangements where women are not
together= 40320-4320 = 36, 000.

It is quite different from the answer obtained by the method taught
here. i.e. 14, 400

ajmalrashid
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THANK GOD THIS LEGEND EXIST!!! THANKS MAN IT HELPED A LOT

JeffersonEspina-tr
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In the first question can't two men stand together? Why are there 11 spaces? when there are only 8 people..? I'm sorry it confused me.

JuiceBoxBoiii
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Thank you very much... Our teacher at school told us that we should get the number of ways of arrangements when their are no restrictions and subtract it from when particular group of people are together in order for us to find the number of ways of arranging them when they are not next to each other...but why didn't this method work on the question in this video?

vincentzulu
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I really don't get how this works. You have 8 possible spaces, but you have made 10. Why could the men not stand next to each other in 1 or 2 instances. The conditions would still be satisfied.

I'm probably being really stupid, but that made no sense to me whatsoever.

DuelWieldingGamers
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Thank you you please also do the opposite I mean if two women are next to each other

singitacatridgenyambi
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thank you so much u got a special place in heaven sir

halah