In how many different ways the letters of the word “MATHEMATICS” can be arranged so that the vowels always come together ? A. 10080 B. 4989600 C. 120960 D. None of these In the word ‘MATHEMATICS’, we treat the vowels AEAI as one letter. Thus, we have MTHMTCS (AEAI).
Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.
Number of ways of arranging
these letters = 8! / (2!)(2!) = 10080.
Now, AEAI has 4 letters in which A occurs 2 times and the rest are different. Number of ways of arranging these letters = 4! / 2! = 12.
Required number of words = (10080 x 12) = 120960. This section covers permutations and combinations. Arranging Objects The number of ways of arranging n unlike objects in a line is n! (pronounced ‘n factorial’). n! = n × (n – 1) × (n – 2) ×…× 3 × 2 × 1 Example How many different ways can the letters P, Q, R, S be arranged? The answer is 4! = 24. This is because there are four spaces to be filled: _, _, _, _ The first space can be filled by any
one of the four letters. The second space can be filled by any of the remaining 3 letters. The third space can be filled by any of the 2 remaining letters and the final space must be filled by the one remaining letter. The total number of possible arrangements is therefore 4 × 3 × 2 × 1 = 4! The number of ways of arranging n objects, of which p of one type are alike, q of a second type are alike, r of a third type are alike, etc is:
n!
.
p! q! r! … Example In how many ways can the letters in the word: STATISTICS be arranged? There are 3 S’s, 2 I’s and 3 T’s in this word, therefore, the number of ways of arranging the letters are: 10!=50 400
3! 2! 3! Rings and Roundabouts - The number of ways of arranging n unlike objects in a ring when clockwise and anticlockwise arrangements are different is (n –
1)!
When clockwise and anti-clockwise arrangements are the same, the number of ways is ½ (n – 1)! Example Ten people go to a party. How many different ways can they be seated? Anti-clockwise and clockwise arrangements are the same. Therefore, the total number of ways is ½ (10-1)! = 181 440 Combinations The number of ways of selecting r objects from n unlike objects is: Example There are 10 balls in a bag numbered from 1 to 10. Three balls are selected at random. How many different ways are there of selecting the three balls? 10C3 =10!=10 × 9 × 8= 120
3! (10 – 3)!3 × 2 × 1 Permutations A
permutation is an ordered arrangement. nPr = n! .
(n – r)! Example In the Match of the Day’s goal of the month competition, you had to pick the top 3 goals out of 10. Since the
order is important, it is the permutation formula which we use. 10P3 =10!
7! = 720 There are therefore 720 different ways of picking the top three goals. Probability The above facts can be used to help solve problems in probability. Example In the National Lottery, 6 numbers are chosen from 49. You
win if the 6 balls you pick match the six balls selected by the machine. What is the probability of winning the National Lottery? The number of ways of choosing 6 numbers from 49 is 49C6 = 13 983 816 . Therefore the probability of winning the lottery is 1/13983816 = 0.000 000 071 5 (3sf), which is about a 1 in 14 million chance.
| 13.
| In how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together?
|
| [A].
| 10080
| | [B].
| 4989600
| | [C].
| 120960
| | [D].
| None of these
|
Answer: Option C Explanation: In the word 'MATHEMATICS', we treat the vowels AEAI as one letter. Thus, we have MTHMTCS (AEAI). Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.
| Number of ways of arranging these letters =
| 8!
| = 10080.
| | (2!)(2!)
|
Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.
| Number of ways of arranging these letters =
| 4!
| = 12.
| | 2!
|
Required number of words = (10080 x 12) = 120960.
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| Arti said: (Jul 28, 2010)
|
| | How we consider 8 letter?
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| Sahithya said: (Sep 10, 2010)
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| | Hi Arti,. After considering vowels as 1 set, remaning consonents and the vowel set => M+T+H+M+T+C+S+(AEAI) => 8 letters {because we hav to consider that the vowels always come together}. I hope you will understand my answer.
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| Ritesh Kumar said: (Mar 4, 2011)
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| | Sir I want to know 10080 and 8/(2!) (2!) how comes.
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| Deepa said: (Jun 1, 2011)
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| | Why dont we use nPr here...as we were using that in last sum. whats the need otherwise where to use nPr..can anybody explain plz ?
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| Tushar said: (Jun 27, 2011)
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| | I want to know 8/(2!) (2!)=10080 how comes.
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| Mini said: (Jul 12, 2011)
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| @ Tushar
After considering vowels as 1 set, remaning consonents and the vowel set => M+T+H+M+T+C+S+(AEAI) = 8 letters {because we hav to consider that the vowels always come together}.
Now 8!/(2!)(2!)
=8*7*6*5*4*3*2*1*/(2*1)*(2*1)=10080
here the (2!) used becoz in the word M+T+H+M+T+C+S+(AEAI)the M occurs twice so one (2!) is for dat and the other (2!) is for T which also occurs twice, while all the other letters including (AEAI)occur only
once.(AEAI) occurs only once bcoz we r cosidering it as one letter.
And for arranging (AEAI)= 4!/2! [bcoz here A occurs twice]
=4*3*2*1/2*1=12
Req. no. of words=10080*12=120960 The formula vich v used in this qestion is
= n!/(P1!)(P2!)....(Pr!) where n is the no. of letters nd P is the no. of occurrence of each letter.
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| Riddhi said: (Jul 24, 2011)
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| | can't we take it as 7p7*4p4.....
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| Usama Zaka said: (Nov 2, 2011)
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| | @Titesh n Tushar 8! meanx (8*7*6*5*4*3*2*1) which gives us 40320 similarly (2!)(2!) gives us 4. So dividing 40320/4 we get 1008. Hope you understand.
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| Krithika said: (Dec 15, 2011)
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| | In the same sum if we are asked to arrange the letters such that the vowels always occur in the same order(i.e- there maybe any no of letters between the vowels, but the order in which the vowels occur in the word should be same).
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| Wolanyo said: (May 14, 2012)
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| | How many ways can we arrange so that the two Ms do not come together?
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| Ram Naik said: (Jun 18, 2013)
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| | @Wolanyo. To find out the number of ways to arrange the so that two Ms do not come together.! Two Ms don't come to gether = Total number of ways - two Ms come together. Total number ways of arranging MATHEMATICS letters = 11!/(2!*2!*2!) = 4989600. Two Ms come together = 10!/(2!*2!) = 907200. Two Ms don't come together = 4989600-907200 = 4082400.
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| Deepa said: (Oct 28, 2014)
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| | Number of ways of arranging these letters = 4!/2! = 2! And than the answer was = 10080*2*1 = 20160.
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| Salman Khan said: (Jan 2, 2015)
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| | Can we use permutation(nPr) method?
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| Vaibhav said: (Jun 14, 2015)
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| | (AEAI) can also have different arrangement like AAEI, AIAE...etc.
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| Raji said: (Aug 24, 2015)
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| | Why don't take 7c4 instead of 7p4?
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| Rohit Soni said: (Sep 24, 2015)
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| | Sir, how have you solved 8!/(2!) (2!) and 4!/2!?
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| Pruthvi said: (Oct 1, 2015)
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| | This answer is wrong we have to multiply by 2 at last which indicates arrange of vowel and other group.
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| Sudama said: (Mar 12, 2016)
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| | Can we use permutation(nPr) method? & also Used Other Methods.
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| Baskar said: (Jun 22, 2016)
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| | In the word 'MATHEMATICS', we treat the vowels AEAI as one letter. Thus, we have MTHMTCS (AEAI). 7! * 4! => 5040 * 24 => 120960. Make it simple!
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| Pranjali said: (Jun 29, 2016)
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| | Don't you think the overall answer should be multiplied by 2? MTHMTCS (AEAI) and (AEAI) MTHMTCS. I am little confused here. Anyone help me to get this.
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| Srav'S said: (Mar 31, 2017)
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| | I am not getting this answer please sir give me a clear explanation.
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| Kerese said: (Apr 28, 2017)
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| | @ALL. We should notice that in the word "mathematics'", m and t occurred twice 8!/2!2!.
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| Ankur said: (Jul 28, 2017)
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| | In the word MATHEMATICS, We treat the vowel as one letter. Thus we have MTHMTCS (AEAI). Now we have t arrange 7 letters, out of which M and T occur twice. So no of arranging these letters = 7!/2!*2! = 1260. Now, AEAI has 4 letters in which A occurs twice. So, no of ways of arranging these letters = 4!/2!=12.
Required no of words= 1260*12=15120.
So answer is none of these.
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| Getnet Tadesse said: (Aug 7, 2017)
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| | MATHEMATICS. # The consonants MTHMTCS are considered as 7 letters. # The four vowels AEAI are considered as one letter since they are supposed to be arranged all together. # The 7 consonant + 1 group of vowels= 8 letters. # If there are no repeated letters in these 8 letters, the total ways of arrangements would be 8 != 40320. # But in the 8 letters we set up above we know M and T are happened to occur twice. Since a letter repeated
twice has two arrangements M and T add up to have 4 repeated arrangements. These 4 repeated arrangements will divide the total 40320 arrangements in to 4. Hence, 40320÷4= 10080. # The AEAI letter will form (4!) 24 arrangements if there is no letter repeated. But A is repeated twice which divide the 24 arrangements by 2. Hence, we do have 12. arrangements. # Finally the total number of ways to arrange is፡. 10080*12=120960 ways.
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| Himanshu said: (Sep 15, 2017)
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| | If vowels never come together then?
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| Rachael said: (Jan 10, 2018)
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| | How can we know that how to separate the vowels?
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| Krishan Senarath said: (Jan 17, 2018)
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| | If four countries are contesting for 5 cups in a competition, how many results can be there?
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| Shree said: (May 21, 2018)
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| | Thanks all for the clear explanations.
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| Gouthami said: (Jun 23, 2018)
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| | My answer is 11!/2!2!2!. Is it is wrong, please explain the answer I have a dought.
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| Chandan said: (Jun 27, 2018)
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| | In the above solution as suggested in the answer section, we are just taking the 4 vowels set at last, but its not said in the question that the vowels set to come together and at last of the word. The words can also be formed as M (AEAI) THMTCS or more which is not considered.
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| Rakesh said: (Aug 22, 2018)
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| | But here are 6 odd spaces and 4 vowels shouldn't we consider that? Please explain to me.
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| Popra Tetseo said: (Oct 7, 2018)
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| | Anyone can please solve this for me? In how many ways PENCIL be arranged so that P and C are next to each other?
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| Maswoad said: (Dec 29, 2018)
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| | @Popra. In the arrangement of letters of word PENCIL, P and C next to each other means P and C always come together and only one possible arrangement of P and C letter that is PC, CP is wrong. Now think of available letters E, N, I, L, [PC]. So possible arrangement = 5! => 120.
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| Prashanth D said: (Jul 6, 2019)
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| | GIVEN WORD = M A T H E M A T I C S = TOTAL = 11. Solution:- VOWEL FROM THE WORD = A A E I -------------------------> (1)
REMAINING LETTERS IN THE WORD = M T H M T C S ------->(2) CONSIDER [ A A E I ] M T H M T C S = 1 + 7 = 8! divided by 2! 2! [becus we have M M and T T]. FROM ------->A A E I we get 4! and divided 2! [becus we have two A A]. 8! 4! DIVIDED BY 2! 2! 2! =======> 120960.
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| Sourabh Sharma said: (Nov 14, 2019)
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| Sir how to find the total arrangements of word ;mathematics, of both A and both M together
AA. MM
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How many ways can MATHEMATICS be arranged so that the vowels come together?
Thus, we have MTHMTCS (AEAI). Number of ways of arranging these letters =8! / ((2!)( 2!)) = 10080.
∴ Required number of words = (10080 x 12) = 120960.
How many ways to arrange the letters in MATHEMATICS if the vowels are always together at the end of word?
120960 ways . Originally Answered: In how many different ways can the letters of the word mathematics be arranged so that the vowels always come together?
How many arrangements are there if all the vowels are together?
Assuming all vowels will be together 15,120 arrangements.
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