Kinh Nghiệm Hướng dẫn In how many ways the letters of the word FAMILY can be arranged when F and Y are always together Mới Nhất
Dương Anh Sơn đang tìm kiếm từ khóa In how many ways the letters of the word FAMILY can be arranged when F and Y are always together được Cập Nhật vào lúc : 2022-09-20 02:40:25 . Với phương châm chia sẻ Kinh Nghiệm về trong nội dung bài viết một cách Chi Tiết Mới Nhất. Nếu sau khi Read nội dung bài viết vẫn ko hiểu thì hoàn toàn có thể lại Comment ở cuối bài để Mình lý giải và hướng dẫn lại nha.Solution:
Nội dung chính- In how many ways can the letters of the word PERMUTATIONS be arranged if the (i) words start with P and end with S, (ii) vowels are all together, (iii) there are always 4 letters between P and S?How many ways can you rearrange the letters in the word family?How many different ways can the letters of the word problem be arranged so that the vowels always come together?How many words can be formed from the letters of the word triangle with t always the beginning and E the end?How many ways can 4 letters be arranged?
We know that the number of arrangements (permutations) that can be made out of n things out of which there are p, q, r, ... number of repetitions = n! / [p! q! r! ...].
In the given word PERMUTATIONS,
No. of T's = 2
Total number of letters = 12.
(i) It is given that words start with P and end with S. So the letters in the first and last positions are fixed. The middle 10 positions have to be filled with the remaining 10 letters among which there are 2 Ts (which are repeated).
No. of words = 10!/2! = 18,14,400.
Click here to see the formula behind it.
(ii) vowels are all together,
We know that there 5 vowels in the given letter. Then it becomes P, R, M, T, T, N, S, (fboxEUAIO), so there are 8 units among which there are 2 Ts'.
Also, vowels can be interchanged within themselves in 5! ways.
Thus, the number of words = 8!/2!× 5! = 24,19,200.
(iii) It is given that there are always 4 letters between P and S. So P and S can take the following positions respectively.
P S 1st 6th 2nd 7th 3rd 8th 4th 9th 5th 10th 6th 11th 7th 12thThere are in total 7 ways in which there are 4 letters between P and S.
If we interchange P and S in the above table, we get 7 more ways of placing P and S.
Thus, total number of ways in which P and S can be placed = 7 + 7 = 14.
Now, the remaining 10 positions have to be filled with the remaining 10 letters among which there are 2 Ts. So
No. of ways for filling the remaining 10 positions = 10!/2!.
Total no. of ways = 14×10!/2! = 2,54,01,600
NCERT Solutions Class 11 Maths Chapter 7 Exercise 7.3 Question 11
In how many ways can the letters of the word PERMUTATIONS be arranged if the (i) words start with P and end with S, (ii) vowels are all together, (iii) there are always 4 letters between P and S?
Summary:
(i) No. of ways = 18,14,400; (ii) If we consider all the 5 vowels as 1 unit, then no. of ways = 24,19,200; (iii) Total no. of ways = 2,54,01,600
2.
In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?
[A]. 360 [B]. 480 [C]. 720 [D]. 5040 [E]. None of theseAnswer: Option C
Explanation:
The word 'LEADING' has 7 different letters.
When the vowels EAI are always together, they can be supposed to form one letter.
Then, we have to arrange the letters LNDG (EAI).
Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.
The vowels (EAI) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120 x 6) = 720.Video Explanation: https://youtu.be/WCEF3iW3H2c
Shoba said: (Sep 4, 2010) Other than vowels there are only 4 letter then how it s possible to get 5!.
Sai said: (Sep 10, 2010) HI SHOBA,.
4 consonants + set of vowels (i. E. , L+N+D+G+ (EAI) ).
We should arrange all these 5. So we get 5!.
I think you understood.
Sachin Kumar said: (Sep 29, 2010) Please make me understand this answer as did not get.
Sri said: (Oct 10, 2010) As vowels are together take (EAI) as single letter i.e. , total no of letters are 5 (L, N, D, G, EAI).
No of ways can arrange these 5 letters are 5! ways.
Now we arranged 5 letters (L, N, D, G, EAI).
Next we have to arrange E, A, I (they may be EAI/EIA/AEI/AIE/IAE/IEA).
All these combinations imply that vowels are together.
So we have to multiply 5! and 3!.
Subbu said: (Jan 6, 2011) 7!=5040
Madhusudan said: (Mar 14, 2011) Could you kindly let me know what is ( ! ).Howe 5! = 120 ?
Sundar said: (Mar 14, 2011) @Madhusudan
5! = 5 Factorial = 1 x 2 x 3 x 4 x 5 = 120
Laxmikanth said: (Mar 26, 2011) When should we take the one or more letters as a single unit and why?
Moaned said: (Apr 8, 2011) I am not getting
Jessie said: (Apr 10, 2011) 7 letter word = LEADING
CONDITION = VOWELS TO BE TOGETHER, HENCE (EAI) TO FOR A WORD
SO NO. OF WORDS = L,(EAI),D,N,G = 5
permuation to arrange 5 letters = nPr= n!/(n-r)!=5!/0!=5!
0! is assumed to be 1!)
EAI can be arranged among each other in = nPr = 3!/(3-3)= 3!
hence 5! x 3! = 120 x 6 = 720
Mayur said: (Apr 15, 2011) How it came like nPr formula and how you have solve it? please let me know.
Bharti said: (Jul 11, 2011) Thanks a lot. I had confusion before your explanations. Thanks a lot.
[email protected] said: (Aug 5, 2011) @mayur.
You just look when ever you open the new exercise there will be availability of basic formulas. If you go through them half of the task would be finished easily. Have a good day buddy.
Siva said: (Oct 16, 2011) Permutations and combinations always make me to confuse much.
How to decide based on the descriptive aptitude question ?
Shiva said: (Oct 18, 2011) Why we have taken 5 (4 + 1) ?
Santosh Kumar Pradhan said: (Oct 27, 2011) Total thành viên 7
out of which 5 consonant and 3 vowels
take 3 vowls as a 1
hen number of consonant will arrange 5! ways
and these 3 vowels will arrange 3! ways
though,5!*3!=720
Bhavik said: (Nov 4, 2011) How to read these nPr ?
Sagar Choudhary said: (Nov 6, 2011) What is this 5! and 3! ?
A.Vamsi Krishna said: (Jan 30, 2012) "!" this implies factorial that means a number is multiplied
like for example take number 5 then its factorial will be
taken as 5*4*3*2*1 and this is equal to 120.
Ncs said: (Feb 2, 2012) Why not 5! + 3! ?
Raj said: (May 21, 2012) Friends, How do you say this question is permutation.
Mahanthesh said: (Jul 30, 2012) Hi guys. As per the question, it s mentioned that all vowels should be together, but it has not been mentioned that it should be EAI. According to me these 3 vowels can be arranged in 3*2=6 ways. And the remaining letters LDNG can be arranged in 4*3*2*1= 24 ways. Can anyone explain me about this please ?
Sandeep said: (Sep 7, 2012) can any one explain ? y 5!*3! & y not like this 5!+3!
Srk said: (Oct 9, 2012) @Mahantesh.
1. There is given a word LEADING in this LDNG (consonents) , EAI (vowels).
2. They asked here vowels always come together and so we should have to take LDNG (EAI).
3. We can take as (4+1) ! i.e. here we have to take vowels as together as 1. So we have a chance of 5!.
4. But with in vowels we have many arrangements i.e 3!
5. Finally 5!*3!
Shafi said: (Oct 11, 2012) @Mahantesh,
You are correct 3! and 4!, because you spitted vowels and consonants,
But to combine them vowels+consonants.
i.e. (4 consonants + 1 set of vowels) = 5!
And (3 vowels 1 set) = 3!
So 5!*3! = 5x4x3x2x1x3x2x1=720.
Adhityasena said: (Feb 9, 2013) I get the answer to be 2*720. Because, since the vowels must come together in the word LEADING, which actually has 7 letters, E and A can be taken as one unit, so now we have L, EA, D, I, N, G to be arranged and which can be done in 6! ways. But among E and A there are 2 arrangements namely EA and AE, So the final answer is 2*720. Am I right !
Chetas said: (Jul 6, 2013) @Adhityasena.
You have not considered a vowel 'I'. 'EAI' is to be taken as one unit.
Sanjay(9776387850) said: (Aug 29, 2013) Hi Guys, first I tell you how can you know is this permutation or combination. Permutation means arrange(row/column) and Combination means selection(group).
i.e. Number and Word are perm. Playing 11 and committee are combi. This is a word so this is perm. You should know the formula that m different objects are alike and n different object are alike if we arrange all the m+n objects such that n objects are always together=(m+1)!*n!
Here n = Vowel = 3 and m = Con. = 4.
So = (4+1)!*3! = 5!*3!.
= 120*6 = 720.
Chithra said: (Nov 5, 2013) As we known very well that * is consider to be AND, + is consider to be OR. Consider the 1st example 7 men and 6 women problem we taken as (7c3*6c2) + (7c4*6c2) + (7c5).
We can also said this as (7c3 AND 6c2) OR (7c4 AND 6c2) OR (7c5).
CONDITION -> 5 people need to select. We should take 3 men from 7 men and 2 women from 6 women or other option is to take 4 men from 7 men and 1 women from 6 women.
That's why we are using * the place of AND, + the place of OR. Similarly, we need to consider both the consonant AND vowel not consonant OR vowel.
So we use 5!*3!.
Taku Mambo Bellnuisemarbel said: (Nov 29, 2013) Please help me with this; whenever I hear of probability that a variable is being selected what should I think of?
Amna Fida said: (Dec 3, 2013) How we know that factorial will use here?
Amnafida said: (Dec 3, 2013) Describe that how we know about here permutation is use?
Jhansi Sri said: (Feb 7, 2014) Please help me quickly why we take 5!*3!, Why we can't take 5!+3!.
Pema said: (Aug 3, 2014) When it comes the question for arrangements, then it is a Permutation Or you all can remember it as keyword "PA" P=permutation and A=arrangement.
Likewise, for combination, it is all for selection purpose, remember keyword as "CS" c=combination,s=selection. Then apply formula for each. Easy.
Baidyanath Jena said: (Nov 7, 2014) When it comes to persons it should be combination.
Samson said: (Nov 17, 2014) God bless you all for your contribution especially you @Jessie for using the formula to break it down well.
Ranjeet said: (Nov 19, 2014) Well I am confused. Somewhere n! is done whereas somewhere (n-1)! is used.
Can someone explain about it?
Sagar said: (Dec 26, 2014) Hi friends.
We know that formula n!=n (n-1) (n-2).....3.2.1. Suppose there n way to choose first element (since there are n elements).
After that there are n-1 ways to choose second element because already we choose one element from n elements that's why we are assuming this way. Similarly n-2 ways to chose the third element..etc it's going like this.
n!=n (n-1).
n!=n (n-1) (n-2) if n>2 or equals 2.
n!=n (n-1) (n-2) (n-3) if n>3 or equals 3.
Hope you understood.
Shantha said: (Apr 30, 2015) Then how we won't take E+A+I+(LDNG) = 4.
Tom said: (May 2, 2015) In what situations we can permutation or combination?
Riya said: (May 8, 2015) @Tom and @Shantha.
Whenever there is a reference to some arrangement it is permutation. Whenever there is a reference to some selection it is combination.
The given question requires us to find the number of ways in which the word LEADING can be arranged with the condition that the vowels (EAI) always be together. Thus we need to apply the concept of permutation.
Here, since the vowels EAI must always be together we consider it as a single word (EAI).
Thus, LDNG (EAI) make a 5 letter word.
It can be arranged in 5p5 ways = 5!ways.
Now, since EAI can arrange itself in 3p3 or 3! ways, the word LDNG (EAI).
Can thus be arranged or PERMUTED in 3!*5! ways = 720 ways.
Spurthy said: (Jun 24, 2015) All the consonants can also be written as a unit?
Ajay said: (Aug 17, 2015) In some cases unit's place, tens place's and hundred's place are used what its mean?
Shrinivas said: (Aug 25, 2015) Hi if two vowels are repeated in same word then will you take it as same or as different?
Vanmathi said: (Sep 1, 2015) In how many ways LEADING be arranged such a way that atleast two vowels always together?
Soumya Sengupta said: (Sep 18, 2015) If there is another vowel 'o' what should be done like in 'outstanding' here a, i, o, u are vowels do we have to consider (aiou) = 1 letter for calculation?
Palak said: (Sep 18, 2015) In this question won't the arrangement of non vowels matter and why?
Raja said: (Sep 19, 2015) Hai @Soumya.
S its 1 letter only. Hope you understand.
Yonatan said: (Sep 24, 2015) I can't understand the answer.
Mukesh said: (Dec 18, 2015) Please anyone help me why repetition is not considered in the current problem?
Eswaru said: (Feb 3, 2022) Hey dude we have to form 7 letter words from LEADING that means we need to use all the letters a time. So no repetition are allowed.
Peter said: (Feb 28, 2022) If least two vowels always together then what will be the answer for LEADING ?
Siva said: (Mar 16, 2022) Other than vowels there are only 4 letter then how it s possible to get 5!.
Brian said: (Jun 2, 2022) Correct me if I'm wrong but,
There are 7 letters:
L E A D I N G
3 vowels and 5 non-vowels.
I got the part where we need to permutate 3 and 5, resulting with 3!*5!. But aren't there also several other positions that these vowels could be positioned?
For Example:
AIELDNG is one factor, another factor is LAIEDNG, another factor is LDAIENG.
As you can see, the positions of the 3 vowels with each other are the same, and the order of non-vowels in the word is also the same, but I only changed the position of the starting point for the vowel permutations, starting from the first position, to the second, to the third. So, I believe that the answer should be 3!*5!*5. Since there are 5 different ways you could represent the same order of vowels in different positions with the same order of non-vowels.
Please do correct me if I'm wrong or if I misunderstood the question
Technothelon said: (Jun 9, 2022) @Brian.
There aren't such kind of ways. Because arrangement is not taken into consideration when we do combinations.
Kirupa Rani D said: (Jun 29, 2022) Can you give solution for this problem? How many words can be formed from the letters of the word 'PACKET', so that the vowels are never together?
Jomson Joy said: (Aug 3, 2022) In PACKET there are 2 vowels. Vowels came 2gether means 4! * 2! = 240.
Total words formed =6! (because of total letters) = 720.
Therefore 720 - 240 = 480.
Gopika S S said: (Aug 3, 2022) A number of permutations of "n" different things are taken "m" specified things always come together is m!* (n-m+1) !
Here the three vowels always come together. So here m is 3. A total number of letters is 7. Substitute we get.
3!* (7 - 3 + 1)! = 3! * 5!
= 6 * 120 =720.
Hope you got it.
Prajwal said: (Aug 17, 2022) Does it have an another method to find the solution?
Rakesh said: (Sep 16, 2022) In leading 5! * 3! ways we can fill all vowels come together.
Otieno said: (Nov 30, 2022) Where has 5! come from? Please explain me.
Xyz said: (Dec 20, 2022) How many letters can be formed from COMBINATION if vowels are kept together?
Please give the solution.
Sowmya said: (Dec 23, 2022) @Xyz.
C,M,B,N,T,N,(O,I,A,I,O)=> C,M,B,T,N,(O,I,A)=> 6! ways.
Vowels alone can be rearranged themselves in 3! ways.
So 6! * 3! = 2160.
Hope this is right.
Navan said: (Feb 2, 2022) I have one question.
How many ways 11players selected from 15 players?
Please give me the answer.
Akshay said: (Mar 31, 2022) @Navan.
15c11 = 15!&div(15-11)! * 11!
Avishek Kadel said: (Apr 21, 2022) In how many ways the letters of word ELEMENT can be arranged so that vowels are always together?
Can anyone solve this?
Ali said: (Apr 26, 2022) @Avishek
No. of vowels is 3Es and the can only be together in this pattern EEE. There is only 1 way of choosing so 1! = 1 * 1 = 1.
We are to consider 3Es as a single alphabet because they are together. So we have L, M, N, T,EEE ie 5 letters now in all.
Now of way of choosing 5 letters = 5! (5 * 4 * 3 * 2 * 1) = 120.
Finally, we multiply 120 * 1 = 120 ie. the number of ways of arranging the letters so that the vowels are always together.
Krishna said: (May 23, 2022) LEADING- vowels together in total 5 positions L, E, A, D, and I.
Ex. For the first position - (EAI) - (remaining 4 lettersLNGD) - 4! * 3!
For total 5 positions - 5*3!*4! -720.
Abhishek said: (Jul 18, 2022) I think it should be 3!*4!
Tanmay said: (Aug 14, 2022) In how many different ways can the letters of the word software be arranged in such a way that the vowels always come together?
Jon said: (Aug 17, 2022) @Tanmay.
SFTWR (OAE)
5 +(1) ! = 6!
OAE can be arranged in 3 ways 3!
6! = 6*5*4*3*2*1 = 720 ways
3! = 6 ways
720*6 = 4320 ways!
Dhana said: (Oct 9, 2022) Word : leading
condition : vowels together
eai-3!
ldng&eai-5!
(ldng)&(eai)-2!
Whether the last condition is valid?
Give explanation.
Nabin Shah said: (Nov 5, 2022) Why do 120 and 6 multiply?
Vivek Sharma said: (Nov 11, 2022) It should be 1440, as the vowels can be after the consonants as well as before the consonants. So 720*2. Am I right?
Heta Vaghasia said: (Nov 12, 2022) Answer will be 72. How 720 came?
Dragonslayer said: (Jan 15, 2022) Friends in the word LEADING.
there are three vowels A,E,I
With AEI you can form 3! arrangements ie 6 arrangements
As we want the vowels to come together they may be placed any of the 5 locations on the word LEADING.
So there are 5! arrangements for this.
5! X 3!=720.
The 3 vowels are considered as a single word to simply this process of counting the number if locations.
Juliana said: (Mar 14, 2022) How do you calculate combinations, can you do an example? Please.
Saravanakumar said: (Apr 6, 2022) LEADING.
Total vowels : E , A, I
remaining : LDNG
assume : EAI we can arrange (EAI)L, L(EAI), D(EAI), N(EAI),G(EAI)
so, total remaining is 4 so, 5!
and possibilities of arranging refer "assume" , 3!
then answer is : 5!*3!
120 * 6=720.
Shivam said: (May 23, 2022) Why 120*6 why not 120+6? Please explain me.
Sankardev said: (Jun 10, 2022) -L-D-N-G.
5!/(5-3)! = 120/2=60.
60*3!=60*6,
=360.
Supriya said: (Jun 22, 2022) LEADING- V-3,(1 unit).
C-4 ,(4+ 1unit)=5!
n-7.
=5!*3!=120*6=720.
Tushar said: (Oct 25, 2022) @Shiva.
It means 5!.
Pooja said: (Dec 31, 2022) 5! means? Explain
Kiran Kumar K said: (Apr 5, 2022) How you get 5(4+1=5) i.e 5!?
Adhi said: (May 22, 2022) Will anyone please explain, how can arrange if one more same vowels had. Eg CORPORATION?
Aniket Arsad said: (Jun 13, 2022) Vowels are EA so _L_D_I_N_G_.
So there is six space we can arrange EA so the answer is 6! = 720.
Pallavi said: (Jun 14, 2022) Why 120*6?
Why not 120+6? Please tell me the reason.
Ameena said: (Aug 2, 2022) Hello,
We have word leading.we have vowels( EAI), first taken as 1 and can be arranged in _L_D_N_G_ that is in 5! Ways, EAI in 3! Ways, and LNDG in 4! ways.
That is 5! * 4! * 3! = 17,280 ways.
Deepanesh said: (Aug 21, 2022) Why we are taking possibilities of vowels too?
Aradhy said: (Sep 22, 2022) @Deepanesh.
We are taking the possibilities of vowel because vowels can also change their position in the given arrangement.
Suraj said: (Oct 28, 2022) Make a queue of vowels should not come together.
Xyz said: (Nov 30, 2022) How many 5 letter words can be formed out of the word NATIONALIST?
How to solve this. Please help!
Keerti Ar said: (Jun 14, 2022) LEADING = 7 letters => 4 alphabets (LNDG) + 1 group (3 vowels) = 5 ! ways.
=>5 * 4 * 3 * 2 * 1 = 120.
Now, the group can be arranged in 3 ways, since there are 3 vowels,
So 3!=6 (3 * 2 * 1).
Hence, 120 * 6 = 720 ways can be arranged.
Dheepak S said: (Jul 26, 2022) I can't understand this, Please anyone help me to get it.
Sanjay Raj G said: (Feb 2, 2022) @All.
Answer is not 720.
Ans is 5040.
L E A D I N G
Total 7!
So 1*2*3*4*5*6*7 = 5040.
Ayush said: (Feb 4, 2022) Here 5! = 5*4*3*2*1 = 120. Post your comments here:
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How many ways can you rearrange the letters in the word family?
It is given that F and Y are always to be together, which means they are one single unit. So the number of letters or units now is 5. The number of different ways in which these letters can be arranged is 240 times. So there are a total of 48 combinations made individually, with a single unit as the beginning.How many different ways can the letters of the word problem be arranged so that the vowels always come together?
Since, the vowels have to be together, we can say that we have to arrange the groups (C), (R), (P), (R), (T), (N) and (OOAIO) among themselves. Considering the objects of the same type, this can be done in. 2 ! = 2520 ways.How many words can be formed from the letters of the word triangle with t always the beginning and E the end?
Therefore, the total number of words formed from the word TRIANGLE is 40,320 out of which 720 words start with T and end with E.How many ways can 4 letters be arranged?
Continuing in this way we have 4! = 4 * 3 * 2 *1 = 24 ways to arrange four letters. Tải thêm tài liệu liên quan đến nội dung bài viết In how many ways the letters of the word FAMILY can be arranged when F and Y are always together