Number of Ways to Arrange "PEPPER"

[@chrisrockscode1202]

This actually helped me out with a algorithm problem I’ve been having, thank you!

[@mrhtutoring]

Glad it helped!

[@amoskowitz0103]

Excellent permutation example. This gentleman is no doubt responsible for many people understanding the why and not just the what! :)

[@nadred5396]

Combination not permutation

[@amoskowitz0103]

@@nadred5396 good point. I stand corrected.

[@XroorX]

When I see a “5” written they way you scribe it, it reminds me of my Korean mom teaching me multiplication tables!

[@jamesmichuki5804]

I love the way this gentleman makes learning maths sooooooo easy!!... almost unbelievable. Wish all teachers emulate this wonderful teacher.

[@PaulbylPaulbyl]

I love this stuff. Brings back wonderful memories. 😊

[@msbunnypaincakes2363]

I like u so much man... I love math and the random math facts that slipped through the cracks when i was in school you always are talking about... Ty

[@mrhtutoring]

Thank you for enjoying the videos! And the compliment. :)

[@DukeWooze]

Permutations and Combinations always tripped me up. This is a great explanation. Thank you.

[@rafaelacastillo5073]

You make math easy to understand. Thank you.

[@mrhtutoring]

You're most welcome

[@wileynicholson1682]

Thanks for making math fun and applicable.

[@mrhtutoring]

You are welcome!

[@aresrizkiwibowo]

Thanks for the videos! They really helped me to understand more or to recap the concepts that I've learned. Short, simple, and easy to understand.

[@mrhtutoring]

Glad to help!

[@AbdulMannan-jx1yc]

This was the easiest to understand thanks

[@antagonisticalex401]

Ah, Permutations and Combinations. Truly one of my favourite topics along with its older brother Probability and (unrelated) Complex Analysis.

[@oilman1758]

You need to teach ALL Math teachers how to teach. You make it so easy to understand and follow. Ill bet everyone in your class will be Engineers and Science majors.

[@zetopr8058]

Howdy, thats a great video! Thank you, i knew it but it was the fact that you counted it and presented it very efficiently and near by the calculation, such that I could follow your thoughts and create my own on it

[@mrhtutoring]

Glad it was helpful!

[@jaredthomas9246]

Iv learned factorials before but I got this so much faster with this explanation. Visual helped better understand what was going on. Thanks 🙏

[@mrhtutoring]

Glad it helped!

[@spmagtfx]

Again. Thanks, I'm reminded of my high school calculus teacher.

[@theperfectidio1]

i didn't know they taught calculus at high school, what school did you go to?

[@gwensimmons_gigi1629]

Yes Perm&Comb!🎉❤

[@lobabaabozahra7725]

You are one of the best channels in the world,thanks, keep up

[@mrhtutoring]

Wow, thanks

[@sofiaknyazeva]

thank you teacher!!

[@satyamsaikrupa]

Sir its a humble request to bring some good level of problems. Thankyou 🙏

[@prormrx]

Now I get what I learnt in the classroom. Thank you so much sir

[@mrhtutoring]

Glad to hear that

[@iWatchYoutube2601]

I just know you have at least some cs background with that math degree. I love the way you work through the process aloud!!

[@Fezezen]

Really nice use of a projector

[@evreixarara4611]

This man must live up to 120 with his mind and ability intact, I might just go back to school and learn math again just because of him here

[@tunistick8044]

For whom asking how the second way was computed. Basically, the rule is: Card(A)=total number of letters whether different or similar!/Π(similar letters!)

[@dragonizeyt]

What is that symbol called?..

[@tunistick8044]

@@dragonizeyt product?

[@kenpinbasar2531]

​@@dragonizeyt Factorial

[@Daniel31216]

​@@dragonizeytIt's capital Pi.

[@TheMSSafi]

genius 😮

[@user-ji6qm3yy5w]

when my math teacher was explaning it I could not understand it however nov I got it

[@FL-gg4dq]

You can tell he is a good teacher. He looks like his arm is so built from whacking kids hands when they get the answer wrong

[@iancoify]

Thank you, I'm about to do my Master's degree and I totally forgot so much of my stats and probabilities since undergrad. Your videos have helped a lot in their short but concise format to bring me back up to speed quickly

[@mrhtutoring]

Glad it was helpful!

[@SWTORLOL87]

The question should state “uniquely arranged.” An argument could be made that since we aren’t going for real words here the letters are just symbols. You can rearrange like symbols even if the outcome looks the same.

[@danstark462]

Lottery tickets and license plates combination. Very interesting.

[@jaydenleow6370]

I thought this vid was solving an English puzzle at first 💀

[@arc8244]

Permutations, my favourite maths topic when i was in 11th grade

[@veot.2869]

I didn't realize (or just) that a pepper was a pepper.

[@it7264]

Kindly solve more problems on permutations and combinations, As they asked almost in every aptitude exam.

[@mrhtutoring]

Will do!

[@SoSkkk]

Come on !!! This is so easy... My recommendation :- solve JEE ADVANCED question

[@shortsshorts3103]

!It can be solved by permutation and combination too.7!/3!×2!

[@sanjamkapoor9843]

I heard that the factorial used at international level is diff than this one?

[@spmagtfx]

Permutations?

[@KanKar23]

I think combinations not permutations

[@electricengine8407]

combinations of permutations

[@jumpkan7008]

permutations not combinations. combinations would be something like How many ways are there to choose 2 letters from the word 'PEPPER'

[@AniqTheGreatKhotle]

Doesn't make sense at all.😢 Why was the R ignored? What are the rules here?

[@alexandersnow5867]

I don’t really understand how this engrains the intuition behind these counting methods. I get that it’s an easy explanation, but I’m missing the jump between the arithmetic and the reasoning behind why we’re allowed to do so.

[@allanjarrettbasi78]

Now this is blowing.

[@Mike-uy7ph]

I remember old times probability 😢

[@al-lg3qd]

My god math is everything

[@hm-zr3uh]

Thanks

[@omjadhav4795]

Combinatorics

[@ongbaangto9884]

Video recommendation: Could you please review questions papers of NEET and JEE(Indian competitive examinations)?

[@dragonizeyt]

Great..

[@benjamingurevitch4097]

I did 6 choose 3 + 3 choose 2

[@AbdisalamFarah391]

I have never seen many people using these kinds of math at work. Most of what we learn end up unused till the day we die.

[@puleraleaooa8110]

I am only giving the counting principle during the 3rd term. 😮

[@farzeenm6205]

This is a basic question in 11th grade ....combination .but yeah good explanation

[@NicBob89]

I learnt the second way in school but I always have trouble remembering how you get to the equation 😂

[@jocelynllamas6600]

I still have questions 😮‍💨

[@eugenelim11]

Combinatorics. The one course I really don't like.

[@zia0571]

🤯

[@P0T00]

Thank you

[@nitjee9027]

❤

[@karrouchkarrouch]

I try to be ok

[@Dubai_life_]

Thanks a lot sir.

[@3dsaulgoodman]

Each "P" should be counted as its own letter, that's wrong

[@mrhtutoring]

If none of the letters repeated, you'd be correct. The reason we divide is because whether you have the 1st "P" or the 2nd "P" or the 3rd "P" in one of the position, for example the first position, it's still the same exact word. I hope this helps.

[@tychophotiou6962]

According to you ppp, ppp ppp and ppp are all different because I rearranged the p's

[@jasonmorris9330]

nPr and nCr commands work faster in the calculator

[@mrhtutoring]

No doubt. But it's good know the math as well.

[@manvibhasker6767]

Sir how can I strong my math ...i am very weak in math

[@mrhtutoring]

Start from the basics. I know it may take time but if the basics is weak, rest will be difficult to learn.

[@manvibhasker6767]

@@mrhtutoringthank you sir ❤️

[@zlol_ssbm]

Yeah but this doesn’t answer how many of those words are actual words, right?

[@mrhtutoring]

Right. That would be an English question. 😀

[@zlol_ssbm]

@@mrhtutoring everything is a math question if you believe hard enough 🙏

[@anujapathak7712]

​@@zlol_ssbm start writing all 60 of them down and count them mate....its not going to take more than a few hours

[@tazguy371]

Or cancel the "6".

[@NotCad]

why divide shouldn’t it be minus

[@Nino-eo8ey]

I replied to a guy asking the same question as you, but it's kind of a long reply, so if you're interested in learning why, this is the explanation: Say we have 3 people, A,B and C and we wanna choose 2 of them. Okay now there are a lot of possibilities, two of them are: AC and CA. You can instantly recognize a problem here, we picked the same 2 guys TWICE, or in other words, this choice is a duplicate. If you take the rest of the possibilities, you'll notice something very interesting: AB and BA, BC and CB. You'll realize that we're always picking the same 2 guys twice (like BC and CB, same 2 guys picked twice) So how do we avoid duplicates? One way to think about this, is that there are 3 DISTINCT CHOICES or in other words, there are 3 choices that are not repeated, which are: AB, BC and AC. These distinct options repeat twice, or mathematically: 3 options * 2 = 6 options. To get rid of the duplicates we just divide ALL THE POSSIBLE OPTIONS (which is 3 factorial) by 2 because again, they repeat twice. (3*2 )/2 = 3 (note: 3*2 is the number of total options) Now that you have the intuition behind dividing, let's check a similar example to the video's one: How many ways can the letters "ABBF" be arranged? Also, exclude any duplicate arrangements. Before excluding duplicates, there are 24 possible arrangements, but I'll only include some of them: ABBF, ABBF, FABB, FABB (Note: Since we have two B's, we can always switch the position of those two B's around giving us ALWAYS a duplicate arrangement.) If I give you the entire list of possible arrangements, you'll just realize "Oh, there are 12 UNIQUE options here, and they're being duplicated" mathematically, this is: 12 unique arrangements * 2 = 24 arrangements. To get rid of the duplicates we just divide by 2: 12*2 / 2 = 12 unique options. If we try to find the possible arrangements for letters like "ABBB", you won't get duplicates, since all unique options repeat (3 factorial) times (ABBB, for example, repeats 6 times, or 3 factorial. Try switching the position of the B's, you'll see that it repeats more than 3. If you wanna know why, I could explain this in another reply, since I don't wanna make this one longer than it is) so we basically divide by 3! to get rid of the repeated possibilities. What to conclude from this: let's call the number of repeated letters N. For every N letters in some word, one possible arrangement will repeat N factorial times! I hope this ridiculously long reply helped you gain some intuition about permutations.

[@NotCad]

@@Nino-eo8ey thank you

[@dineshpalanii]

60

[@lukocius]

Nah man, numerator explained fine, denumerator is not clear at all, repeated 3 times, so 3×2×1, also divide by 2?

[@ottohammons6232]

why is it assumed that letters cant repeat??

[@Daniel31216]

Because you're arranging the letters you have. It's similar to arranging people, you can't have one person in multiple places.

[@kuijaye]

it seems nobody can really explain this topic well.. why dividing by 3! and 2!? didnt get it

[@kuijaye]

@@roger-yt-192 that's what he said but for some reason didn't explain in detail

[@roger-yt-192]

@@kuijaye Oh lmao, sorry then

[@tychophotiou6962]

To get it. Try writing down all the arrangements of ABCD, and then repeat this process for AABB.

[@kuijaye]

@@tychophotiou6962thanks. this helps building intuition I guess

[@Nino-eo8ey]

It's a pretty simple concept, but it's harder to explain through text rather than visually. I'll try my absolute best to explain to you why that's the case. This might be a long read, but I made sure that everything is easy to understand, if not, just reply which part got you confused and I'll be sure to reply! Here we go: Say we have 3 people, A,B and C and we wanna choose 2 of them. Okay now there are a lot of possibilities, two of them are: AC and CA. You can instantly recognize a problem here, we picked the same 2 guys TWICE, or in other words, this choice is a duplicate. If you take the rest of the possibilities, you'll notice something very interesting: AB and BA, BC and CB. You'll realize that we're always picking the same 2 guys twice (like BC and CB, same 2 guys picked twice) So how do we avoid duplicates? One way to think about this, is that there are 3 DISTINCT CHOICES or in other words, there are 3 choices that are not repeated, which are: AB, BC and AC. These distinct options repeat twice, or mathematically: 3 options * 2 = 6 options. To get rid of the duplicates we just divide ALL THE POSSIBLE OPTIONS (which is 3 factorial) by 2 because again, they repeat twice. (3*2 )/2 = 3 (note: 3*2 is the number of total options) Now that you have the intuition behind dividing, let's check a similar example to the video's one: How many ways can the letters "ABBF" be arranged? Also, exclude any duplicate arrangements. Before excluding duplicates, there are 24 possible arrangements, but I'll only include some of them: ABBF, ABBF, FABB, FABB (Note: Since we have two B's, we can always switch the position of those two B's around giving us ALWAYS a duplicate arrangement.) If I give you the entire list of possible arrangements, you'll just realize "Oh, there are 12 UNIQUE options here, and they're being duplicated" mathematically, this is: 12 unique arrangements * 2 = 24 arrangements. To get rid of the duplicates we just divide by 2: 12*2 / 2 = 12 unique options. If we try to find the possible arrangements for letters like "ABBB", you won't get duplicates, since all unique options repeat (3 factorial) times (ABBB, for example, repeats 6 times, or 3 factorial. Try switching the position of the B's, you'll see that it repeats more than 3. If you wanna know why, I could explain this in another reply, since I don't wanna make this one longer than it is) so we basically divide by 3! to get rid of the repeated possibilities. What to conclude from this: let's call the number of repeated letters N. For every N letters in some word, one possible arrangement will repeat N factorial times! I hope this ridiculously long reply helped you gain some intuition about permutations.

[@stevehendon4076]

The correct answer is 720

[@Biscuit-Loves-All]

I tried with Fourier and I got 2520

[@ravikrishna6273]

I can't understand, Could you please explain briefly sir

[@mrhtutoring]

It may be difficult to understand because I had to explain in 60 seconds for KZbin shorts. I'll make a more thorough video in a regular video soon

[@user-pq9io8td5i]

Huh?🤓