Permutaion and combination1.If three persons enter a bus in which there are 10 vacant seats, find in howmany ways they can sit? [720]2.How many plates of vehicles consisting of 4 different digits can be made out of the integers 4,5,6,7,8,9?[360]3.In how many ways can four boys and three girls be seated in a row containing seven seats.
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4.In how many ways can eight people be seated in a row of eight seats so that two particular persons are always together?
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5.Six different books are arranged on a self. Find the number of different ways in which the two particular books are a]
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6.In how many ways can four red beads, five white beads and three blue beads be arranged in a row?7.In how many ways can the letters of the following words be arranged?a]ELEMENT[840]c]NOTATION[6720]b]MATHEMATICSd]MISSISSIPPI8.How many numbers of 6 digits can be formed with the digits 2,3,2,0,3,3?[50]9.In how many ways can 4 art students and 4 science students be arranged in circular table if a] they sit anywhere. b] theysit alternate[5040,144]10.In how many ways can eight people be seated in a round table if two people insiston sitting next to each other?[1440]11.In how many ways can seven different beads be made into a braclet?[360]12.In how many ways can 4 letters be posted in six letter boxes?13.How many even numbers of 3 digits can be formed when repetition of digits is allowed?14.In how many ways can 3 prizes be distributed among 4 students so that each student may receive any number of prizes?15.In how many ways can the letters of the word ‘MONDAY’ be arranged? How many of these arrangements do not beginwith the letter M? How many begin with M and do not end in Y?[600,96]16.Show that the number of ways in which the letters of the word
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Mohammed
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In how many different ways can the letters of the word ARRANGE be arranged? If the two 'R's do not occur together, then how many arrangements can be made? if besides the two R's the two A's also do not occur together, then how many permutations will be obtained?
Click here👆to get an answer to your question ✍️ In how many different ways can the letters of the word ARRANGE be arranged? If the two 'R's do not occur together, then how many arrangements can be made? if besides the two R's the two A's also do not occur together, then how many permutations will be obtained?
Question
In how many different ways can the letters of the word ARRANGE be arranged? If the two 'R's do not occur together, then how many arrangements can be made? if besides the two R's the two A's also do not occur together, then how many permutations will be obtained?
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The word "ARRANGE" can be arranged in
2!×2! 7! = 4 5040 =1260 ways.
For the two R's do occur together, let us make a group of R's taking from "ARRANGE" and permute them.
Then the number of ways =
2! 6! =360.
The number ways to arrange "ARRANGE", where two "R's" will not occur together is =1260−360=900.
Also in the same way, the number of ways where two "A's" are together is 360.
The number of ways where two "A's" and two "R's" are together is 5!=120.
The number of ways where neither two "A's" nor two "R's" are together is =1260−[360+360]+120=660.
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The number of ways in which the letters of the word ARRANGE be arranged so that [i] the two R's are never to
The letters of word ARRANGE can be rewritten as A R N G E A R So we have 2 A's and 2 R's , and total 7 letters. [i] Total number of words is [7!]/[2!2!]=1260. The number of words in which 2 R's are together [consider [R R] as one unit] is 6!//2!. e.g., The number of words in which 2 R's are together [consider [R R] as one unit] is 6!//2!. e.g., [R R],A,A,N,G,E Note that permutations of R R give nothing extra. Therefore, the number of words in which the two R's are never together is [7!]/[2!2!]-[6!]/[2!]=900 [ii] The number of words in which both A's are together is 6!//2!=360, e.g., [A A],R,R,N,G,E The number of words in which both A's and both R's are together is 5!=120, e.g., [A A], [R R], N,G,E Therefore, the number of words in which both A's are together but the two R's are not together is 360-120=240. [iii] There are in all 900 words in each of which the two R's are never together. Consider any such word. Either the two A's are together or the two A's not together. But the number of all such arrangements in which the two A's are together is 240. Hence, the number of all such arrangements in which the two A's not together is 900-240=660.
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Permutation And Combination
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The number of ways in which the letters of the word ARRANGE be arranged so that [i] the two R's are never together, [ii] the two A's are together but not two R's. [iii] neither two A's nor two R's are together.
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The letters of word ARRANGE can be rewritten as
A R N G E A R
So we have 2 A's and 2 R's , and total 7 letters.
[i] Total number of words is
7! 2!2! =1260 7!2!2!=1260 .
The number of words in which 2 R's are together [consider [R R] as one unit] is
6!/2! 6!/2! . e.g.,
The number of words in which 2 R's are together [consider [R R] as one unit] is
6!/2! 6!/2! . e.g., [R R],A,A,N,G,E
Note that permutations of R R give nothing extra. Therefore, the number of words in which the two R's are never together is
7! 2!2! − 6! 2! =900 7!2!2!-6!2!=900
[ii] The number of words in which both A's are together is
6!/2≠360 6!/2≠360 , e.g., [A A],R,R,N,G,E
The number of words in which both A's and both R's are together is 5!=120, e.g.,
[A A], [R R], N,G,E
Therefore, the number of words in which both A's are together but the two R's are not together is 360-120=240.
[iii] There are in all 900 words in each of which the two R's are never together. Consider any such word. Either the two A's are together or the two A's not together. But the number of all such arrangements in which the two A's are together is 240. Hence, the number of all such arrangements in which the two A's not together is 900-240=660.
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Very Important Questions
उपरोक्त प्रश्न में यदि बल F क्षैतिज से
θ θ कोण नीचे हो, तो θ θ
का मान बढ़ाने पर सीमांत व गतिज घर्षण बल पर क्या प्रभाव पड़ेगा ?
किसी कार के टायरों में पूरी हवा भरने से हम पैट्रोल कैसे बचाते हैं ?
एक मोटर कार सीधी सड़क पर एकसमान चाल u से गतिमान है । सड़क तथा टायरों के बीच घर्षण गुणांक
μ μ
है । यदि कार का इंजन बन्द कर दिया जाये तो कार - [i] कितनी दूर जाकर, [ii] कितने समय बाद रुक जायेगी ?
किसी गुटके व क्षैतिज तल के बीच सीमांत घर्षण 20 न्यूटन तथा गतिज घर्षण 15 न्यूटन है । यदि इसे क्षैतिज बल F से खींचा जाये तो घर्षण बल क्या होगा जब -
[i] F=10 F=10 न्यूटन, [ii] F=20 F=20 न्यूटन, [iii] F=30 F=30 न्यूटन
m द्रव्यमान का गुटका
θ θ
कोण के आनत बल पर है । गुटके पर घर्षण बल क्या होगा यदि [i] गुटका स्थिर है [ii] गुटका गतिमान है।
2 किग्रा के एक गुटके को क्षैतिज फर्श पर रखकर 2.5 न्यूटन का क्षैतिज बल लगाया जाता है । यदि गुटके व फर्श के बीच स्थैतिक घर्षण गुणांक 0.40 हो तो इनके बीच घर्षण बल ज्ञात कीजिये ।
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In how many ways can the letters of the word “arrange” be arranged so that neither two As nor two Rs come together?
Answer [1 of 7]: Edit: my original answer was wrong because it failed to account for the fact that there are two aa in “arrange”. Fixed That is one example that is easily handled by using complements: calculate the total number of ways you can arrange the letters; subtract the number of ways yo...
In how many ways can the letters of the word “arrange” be arranged so that neither two As nor two Rs come together?
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7 Answers Aishwarya Singh
, studied at Loreto Convent, Lucknow
Answered 4 years ago
Total no of arrangements possible=7!/2!2!=1260
Total no of arrangements in which Rs are together=6!/2!=360
Total no of arrangements in which 2 As are together=6!/2!=360
Total no of arrangements in which 2 As as well as 2 R’so are together= 5!=120
Therefore total no. of arrangements in which neither 2 As nor 2 Rs are together= 1260-360-360+120=660
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Related questions More answers below
In how many ways can the letters of the word 'ARRANGE' be arranged? How many of these arrangements are in which [1] two R's come together [2] the two R's do not come together [3] the two R's and the two A's come together?
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How many ways can the letters of the word 'arrange' be arranged so that the two As are together but not the Rs?
Gilbert Duy Pham Doan
, Did Combinatorics with Emille Lawrence
Answered 4 years ago · Author has 15.8K answers and 9.2M answer views
The word is “arrange”. With letters, it has 2 as, 2 rs, 1 n, 1 g, 1 e. It has a total of 7 of any letter. We may not distinguish repetitions. Then
# possible = 7! / [2!*2!*1!*1!*1!] = 7! / [2!*2!]
# favorable = # possible - 5!/[2!*2!]
The portion may be justified as follows. Let 2 as be together, then count them as 1 A. Let 2 rs be together, then count them as 1 R. Keep 1 n, 1 g, and 1 e the same. Then some word has a total of 5 letters. It has 5! distinct rearrangements. And requires divide out repetitions.
It may appear trivial yet not. Since the complement of neither 2 as nor 2 rs is both 2 as
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Luís Sequeira
, PhD Mathematics, University of Lisbon [2001]
Updated 2 years ago · Author has 1.3K answers and 595.5K answer views
Originally Answered: How many ways do letters of the word ‘arrange’ be arranged in such that 2 Rs won't come together?
Edit: my original answer was wrong because it failed to account for the fact that there are two aa in “arrange”. Fixed
That is one example that is easily handled by using complements:
calculate the total number of ways you can arrange the letters; subtract the number of ways you can arrange them with two RR together.
You can arrange seven letters in 7! ways, but since permuting the two R does not change anything you have to divide by 2, and since there also two AA, you have to divide by 2 again.
Then to count the number of always haveing the two RR together, just take them as if one letter, and th
Isaac Sofair
, Math Enthusiast and Software Engineer
Answered 2 years ago · Author has 136 answers and 74K answer views
There are 2 r’s and 2 a’s. So the total number of permutations is 7!/2^2. If the 2 r’s are kept together, then the number of permutations is 6!/2. Hence the number of permutations such that the 2 r’s don’t come together is
7!/4 - 6!/2 = 900.
If you stipulate that both r’s and both a’s don’t come together, then the number of permutations becomes by the principle of Inclusion and Exclusion
7!/4 - 2[6!/2] + 5! = 660.
Note that I have updated my second result.
6.3K viewsView upvotes
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Lawrence Stewart
, CTO at Serissa Research [2001-present]
Answered 3 years ago · Author has 7.6K answers and 8.3M answer views
This is easy to figure out in Python:
>>> import itertools
>>> l = ["".join[x] for x in itertools.permutations['arrange']]
>>> len[l] 5040
>>> l2 = [x for x in l if x.find['rr'] == -1 and x.find['aa'] == -1]
>>> len[l2] 2640
Line 1 says we are going to use the itertools package, which contains permutations.
Line 2 asks for all permutations of ‘arrange’
Line 3 and 4 say there are 5040 of them
Line 5 asks for all the permutations that do not contain ‘aa’ or ‘rr’
Lines 6 and 7 say there are 2640 of them
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Robert Barron
, Retired professor of physics
Answered 4 years ago
Think of the two a’s and the two n’s as individual “letters”. Then there are 5! ways to arrange these “letters” with the other three actual letters. This means that there are 5! ways they actually do come together. The number of arrangement of seven objects is 7! if all the objects were distinct. But the two a’s and two r’s are not distinct and can be permuted in 2! ways. So there are 7!/[2!2!] actual distinct arrangements. Therefore the answer is 7!/[2!2!] - 5! which equals 1140.
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