Permutating YOUTUBE

[@okress]

Thank you for this video ❤. Just a short question: I'm not sure if I understood you right. If we have the word YouTute I think the answer would be (7!)/(2*2) because of the double t and the double u. Is that right? And if we have a word like Yoututu. Would the solution be (7!)/(3*2) because of the three u and the two t? Thank you for your great content. You are such a sympathetic and well-explaining man ❤

[@PrimeNewtons]

7!/3! For your case. 7!/2! For my case.

[@okress]

@@PrimeNewtons thank you. Another question: wouldn't 7!/3! be for a word like Youtubu so that you have three times the u. But if there are also like two t, for example Yoututu would it be 7!/3!*2! ? Sorry if this question is repeating my first but I'm not sure if I understood it right.

[@PrimeNewtons]

You are right.

[@eduardoteixeira869]

Sorry, I believe the answer should be 7!/2!2! for your case. 7!/2! for my case.@@PrimeNewtons

[@d_0364]

for aabbccdef, it will be 9!/(2!*2!*2!). you can generalize it further. for aaabbbccc it will be 9!/(3!*3!*3!)

[@RavenMobile]

I knew the math for digits (decimal, hexadecimal, octal, etc.), but had no idea how it worked for rearrangements. Thanks for the clear video.

[@Rahullegend990]

😂😂😂😂 meanwhile 11th students it is first question of PNC

[@pajthefaj9396]

Let's go, my guy got to 100k subscribers. Been here since 5k. Hope you keep this rapid growth

[@PrimeNewtons]

Thank you! We are still moving forward.

[@MixDSweep]

Love this! I would like more videos about combinations!

[@GolfTunado]

I was subscribing and the sub's number went to 100k. Congrats to 100k subs man... I'm glad KZbin recommended your channel for me!

[@joeljohn839]

I did this completely in my head yay congrats on 100k

[@mervmartin2112]

The permutation tree would be interesting also. You showed 7 letters taken 7 at a time. 7 taken 6 at a time, 7 taken 5 at a time, 7 taken 4 at a time, and so on ? And of course there's combinations too. You present your topic well. I'd love to see you do all topics in the realm of permutations and combinations.

[@amiraliazimi6355]

Thank you love these videos please upload more permutations and combinations videos

[@kujmous]

Your explanation makes sense. My first answer was (6×6!), but I made the following mistake. Removing the U, and starting with only the letters YOUTBE, I had 6!. Then I saw that there are seven total possible ways to place the remaining U, but that placing it just before or just after the existing U would be a duplication... leaving only six unique ways. Here is an example of my oversight. Inserting the U third into TOBEUY and inserting it sixth into TOUBEY are not unique results.

[@wavingbuddy3535]

if we have a set {a1,a2,…,an}, for each group of k indistinguishable elements, we need to divide our initial result of n! permutations by k! Say you have the set {a,a,a,b,b,c,d}, you would do 7!/( 3! •2!) = 420 unique permutations

[@d_0364]

Congratz for 100k sirrrr. . .for past 3 days it was like 99.3 to 99.7. Today it is finally rounded off, now I can finally rest ;). Best of Luck for the future sir. P.S. I would love to see you explaining and solving limits, integration, conic sections etc.

[@wesleybilly8097]

License plate idea sounds fun.

[@BurrritoYT]

I immediately figured out that this would be 2520, still good video

[@RandomGuyEdit]

Congratulations Prime Newtons for your 100K subscribers ... Thank you for these amazing videos I will be thankful to you if you might teach Diophantine Equations .. It will be really helpful ...

[@bombergame8636]

Uoteyub fellas Jokes aside, wish you an amazing start of 2024 man, you truly deserve it

[@foisalmahdi]

Hey! Congratulations for hitting 100k subscribers.

[@vincentaace4939]

I learn combinatorics on my own. And i've seen such a problem solved in a pdf book about set theory and combinatorics. There was also a formula: n!/k1! * k2! * k3! * km!, where "n" stands for the general number of letters and "k" is a number of identical letters. That was called arrangement with repeatings. (Quite surprisingly for me to see a familiar problem here, which i can solve)

[@todddean7722]

I would love to see more combinatorial theory and optimization!

[@5Stars49]

Interesting,2520 is smallest number which is divisible by 1,2,3,4,5,6,7,8,9,and 10.😊😊😅

[@RayMyName]

def not by 11 bc the alternating digit sum is not divisible by 11

[@5Stars49]

@@RayMyName Sorry you are right..This is not divisible by 11

[@person6098]

Never stop learning ❤❤

[@JASONKINGMATHK]

Letter. Is 6! Over 2! 2!

[@legitx420]

Congratulations 🎈 100k sub

[@igeorgiusi]

I think you need to specify that you divide by 2! and not just 2, as it is fine in this case but for higher numbers of repeated elements it won't work if we don't divide by the factorial

[@patricklaenen3468]

Wanted to give the same comment. Imagine you have 7 identical letters, you should divide by 7! to get the answer of only one possible arrangement and not divide by 7 7!/2! is indeed the correct answer which in this case by accident gives the same result as 7!/2

[@maikrentsch6392]

You must divide by 2! In Case of 3 u's you divide by 6 and not 3 (Multinomialcoefficient)

[@jacobgoldman5780]

Another way to think about this since U is the only repeated letter: how many ways to arrange Y,O,T,B,E in 7 spaces and the 2 Us will fill in the remaining spaces in 1 way. Then we get 7 options, times 6 options, times 5 options, times 4 options, times 3 options for the 5th letter so 7x6x5x4x3=2520.

[@Unnayan-jb5ft]

As an iit jee aspirant this is the most basic question of permutation

[@Arya3-6-9]

We actually divide by factorials of numbers of repetitions, for example if there were 3 U instead of 2, the answer would be 7!/3! , and if another element is repeated as well, for example there are 2 T and 3 U, then It would be 7!/(3!×2!), 3! because of 3 U and 2! because of 2 T

[@yankoly]

Thanks you very much sir Please do more videos on PROBABILITY

[@bawatabetando6902]

Thank you forcthis eye opener!!!😊

[@learnwithdk5533]

Yes please continnue you are best i love you sir❤❤❤❤❤

[@OnlineMathsAcademyTV]

Hitting 100k is huge. 200k loading.....

[@PrimeNewtons]

Thank you. You will get there too, even soon.

[@RyanLewis-Johnson-wq6xs]

In how many ways can the letters of the word KZbin be arranged? 2,520 final answer

[@sulimanibra5332]

هايل..جميل👍

[@Aussiesnrg]

That was a very nice way of explaining it.

[@mohammedtallouk349]

Hi Mr. Newton! I wish you a wonderful New Year. Thank you for all the knowledge you share. I like the way you explain things. You are very organized. Thank you so much. 👍.......... .....🌿🌴🌹.

[@ReyBeastGamer]

This video gonna blow up and i mean it

[@n3l3sh]

So the number of arrangements of the Letter ''Banana" 6!/(3!x2!) ???

[@I_Am_Tomas]

Yes

[@m.h.6470]

Solution: It highly depends on whether you want the two "U"s to be separate entities or two instances of the same entity. In other words: Is there a U₁ and a U₂ or is there two U's? In the first case, the answer is just 7! (=5040), as you pick one of the 7 letters first, then one of the remaining 6 second, and so on. In the second case, the answer is 7!/2! (=2520). You basically do the same thing as in the first case, but you discard all duplicate results that appear because of the double U's.

[@mikuculus3720]

Didn’t know about the divide by 2 part. What if the y was also repeated would it be divisible by 4?

[@stephenlesliebrown5959]

Yes, because it's then 7!/(2!2!).

[@mikuculus3720]

@@stephenlesliebrown5959 thanks

[@MarCamus]

I was a little bit confused when you said that we need to divide 7! by 2. And then I told myself what would happen if I had the word LOL, for instance, - 3! is wrong since there are only 3 options to rearrange the letters: LOL,OLL, and LLO. There's no reason to count the letter L twice because the order doesn't matter.

[@stephenlesliebrown5959]

We can keep using the same pattern. Three letters for three boxes and two of letters are the same calls for calculating 3!/2! = 3*2 / 2 = 3.

[@atharvabhardwaj3830]

Please do a video on combinations

[@the_llaw]

CONGRATS FOR 100K

[@surcomsys]

Permutations SHOULD NOT be divided by the number of similar elements, but their factorial. 7! / 2!, or 7! / 3! and so on. Not 7!/2, 7!/3

[@jim2376]

Factorial of the number of letters divided by the factorial of number of letters that repeat. 7!/2! = 2520. Mississippi: 11!/(4! 4! 2!) = 34,650

[@stephenlesliebrown5959]

Right, but please check the arithmetic on the final calculation. I got 34,650 🙂

[@alexnoussi]

Impressive.

[@jim2376]

@@stephenlesliebrown5959 Typo, my bad. Thanks for the correction. I've corrected it.

[@dougaugustine4075]

Yes!! This subject has been a terrible source of bewilderment for me. I once tried, without success, to figure out the total number of ways the back row of chess pieces can be arranged. There are two rooks, two knights and two bishops. The total number of pieces is eight. Is it 8! / 2x2x2?

[@PrimeNewtons]

Unless you you can uniquely identify each piece, it can't be 8!. It has to be 8!/( 2!2!2!)

[@tapansaharia7115]

Sir,,, Please explain what is this --- 225×10 all over 3!>!861 I have no idea what this is.. I am a student of class 8th

[@natureboymoments]

Maximum respect sir

[@Amr_patman]

How to find range of f(x)=absolute1/x Algebraically

[@agamehdialiyev5801]

Actually since U is repeated 2 times then denominator should be 2! not just 2. For instance if you have 3 characters of word like "AAA" the how many combination would you have 3!/3!=1

[@alexnoussi]

My take is "never gamble".

[@stephenlesliebrown5959]

That's fine, but it's good to gamble if the odds are truly in your favor. The difficulty is knowing the true odds. Life (existence) is a gamble. The world is a casino. Best wishes to all 🙂

[@SHOYEBAHMED-q3b]

Can anyone tell me that how many ways to arranged-“GONITZOGGO”

[@PrimeNewtons]

10!/(3!*3!). There are 3 Gs and 3 Os

[@SHOYEBAHMED-q3b]

@@PrimeNewtons Thank you.

[@kevinbush4300]

Oh gosh! Can't believe I made a mistake here... Okay, what I did was I noticed the duplicate 'U' from the outset, so decided that for the first letter I had 6 choices (not 7). Similarly 5 options for the second letter. So in the end, my answer was 6! I can see how your logic is correct, but can't see why mine was wrong. Help!!

[@okress]

I'm not sure but maybe because you acted like you only have six letters but there are seven places to fill. Maybe this helps a little bit but I'm not an expert

[@killianobrien2007]

If letter 1 is u then letter 2 does not have 5 options

[@_realnoob2050]

you forced u's to be filled at last

[@kevinbush4300]

@killianobrien2007 yeah, but before you even start, letter 1 does not have 7 options. Well, seven options exist; but two are equivalent.

[@kevinbush4300]

@okress but the last position is fixed anyway, by the previous selections... So 7! = 7 x 6 x 5 x 4 x 3 x 2 (but by convention we also write "x 1")

[@lightyagami9486]

please continue this topic

[@roythoppilchacko8358]

How did u do that 2520 that fast sir

[@PrimeNewtons]

Haha. Remember, I prepared the script myself.

[@roythoppilchacko8358]

It's is least number divisible by 1-10

[@roythoppilchacko8358]

@@PrimeNewtons my mathematics teacher has same thought of your sir

[@Hawk1996-f4w]

Prof . You did wrong If we have three litters [xxx] Three factorial is 6 Three litters repeated 6/3=2 But we have only one way ( xxx )

[@PrimeNewtons]

That would be 3! Not 3 in the denominator

[@Hawk1996-f4w]

@@PrimeNewtons oh ok thank you

[@Riskw-mk1lo]

letters*

[@ibrahimali3192]

my guess is 2520

[@xerra7293]

I BET IM THE FIRST TO SAY CONGRATS ON 100K

[@PrimeNewtons]

You are! Thank youuuuu!!

[@sinichitaniyama]

handsome man

[@GPSPYHGPSPYH-ds7gu]

7 is already specific number is the whole Universe PAZA M C69AoneA

[@Rahullegend990]

Jee students be like 😅😅😅

[@withershin]

U+0055 and U+0075 are two different things. Please learn some basic computer math. It's been around for decades. I am assuming UTF-8. This math is old. Older than me.

[@uncreatedlogos]

7!/2 = 2520

[@mouthiknaradas962]

7!/2!

[@sudeepranganath5975]

nPr, n as 7 and r as 5