Correct Answer:
Description for Correct answer:
In the word CANDIDATE, letters C, N, D, D. T are consonants and A, I. A, E are vowels.
We have to arrange C. N. D. D. T [A l A E] in which 'D' comes twice and A comes twice. Therefore, Number of arrangements
= \[ \Large \frac{6! \times 4!}{2!2!} \]
= \[ \Large \frac{6 \times 5 \times 4 \times 3 \times 2 \times 4 \times 3 \times 2}{2 \times 2} \]
= 4320
Part of solved Permutation and combination questions and answers : >> Aptitude >> Permutation and combination
Mohammed
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get how many arrangements of the letters of the word keyboard can be made if the vowels are to occupy only odd places in the word arrangement? from screen.
How many arrangements of the letters of the word "KEYBOARD" can be made if the vowels are to occupy only odd places in the word arrangement?. Created by nilanjana2509. Math
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How many arrangements of the letters of the word 'BENGALI' can be madei] If the vowels are never togetherii] If the vowels are to occupy only odd places
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In how many ways can the letters KEYBOARD be arranged, so that the vowels are surrounded by vowels?
Answer: The answer is zero. In any permutation of the letters in “KEYBOARD”, there will always be at least one constant next to a vowel and therefore that vowel is not surrounded by vowels. Therefore there are zero ways can the letters KEYBOARD be arranged, so that the vowels are surrounded by vo...
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Mohammed
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Quantitative Aptitude
Placement-Preparation updated on Aug 2022
A permutation is an arrangement of all or part of a set of objects, with regard to the order of the arrangement. The combination is a way of selecting items from a collection, such that [unlike permutations] the order of selection does not matter. Permutation and combination is all about counting and arrangements made from a certain group of data.
Quantitative Aptitude
Quantitative Aptitude Permutation and Combination-Exercise Questions
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More Related Content
1. Evaluate 50! 47! a. 102500 b. 112584 c. 117600 d. 118450
2. Find the value of 85P3 .
a. 565350 b. 595650 c. 535950 d. 565350
3. Find the value of [20C18]*[20C20]
a. 400 b. 380 c. 360 d. 350
4. How many words with or without meaning, can be formed by using all the letters of the word, ‘ORANGE’, using each letter exactly once?
a. 700 b. 720 c. 750 d. 800
5. There are 28 stations between Ernakulam and Chennai. How many second class tickets have to be printed, so that a passenger can travel from one station to any other station?
a. 800 b. 820 c. 850 d. 870
6. In how many ways can the letters of the word, ‘TECHNOLOGY’ be arranged?
a. 1804400 b. 1814400 c. 1714400 d. 1704400
7. A bag contains 2 yellow balls, 3 white balls and 5 red balls. In how many ways can two balls be drawn from the bag?
a. 2C2 b.10C2 c.8C2 d.5C2
8. In how many ways can the letters of the word, ‘LANGUAGE’ be arranged in such a way that the vowels always come together?
a. 600 b. 700 c. 720 d. 750
9. In how many ways can the letters of the word, ‘KEYBOARD’ be arranged in such a way that the vowels always come together?
a. 4250 b. 4520 c. 4320 d. 4230
10. In how many ways can a team 16 be chosen out of a batch of 20 players?
a. 4845 b. 6852 c. 3125 d. 5846
11. How many ways can the letters of the word, ‘MACHINE’ be arranged so that the vowels may occupy only the odd positions?
a. 210 b. 576 c.144 d. 456
12. From a group of 5men and 4 women, 3 persons are to be selected to form a committee so that at least 2 men are there are on the committee. In how many ways can it be done?
a. 20 b. 50 c. 65 d. 86
13. In how many ways can a committee consisting of 4 men and 5 women be formed from a group of 7 men and 9 women?
a. 7C4 9C5 b. 4C7 5C9 c. 7C5 9C4 d. 9C4 7C5
14. In how many ways can 5 boys and 3 girls sit around a table in such a way that no two girls sit together?
a. 1000 b. 1400 c.1440 d. 1800
Directions for questions 15 to 16: Refer the data below and answer the questions below:
A letter lock has 3 rings each containing 6 letters.
15. What is the maximum number of false trials that can be made before the lock is opened?
a. 3*26C6 b. [26C6]3 c. 26C6. 3! d. 215
16. How many such three letter passwords can exist?
a. 216 b. 26C6 *3 c. [26C6]3 d. [26C6]3 *63
17. How many different words can be formed from the word DAUGHTER so that ending and beginning letters are consonants?
a. 7200 b. 14400 c. 360 d. 1440
18. Out of 6 consonants and 3 vowels, how many words of 4 consonants and 2 vowels can be formed?
a. 1050 b. 25200 c. 32400 d. 5800
19. A box contains 3 white balls, 4 black balls and 5 yellow balls. In how many ways can 4 balls be drawn from the box, if at least one yellow ball is to be included in the draw?
a. 652 b. 547 c.425 d. 356
20. In how many ways can 22 books on English and 20 books on Hindi be placed in a row on a shelf so that two books on Hindi may not be together?
a. 4586 b. 5896 c. 2415 d. 1771
Answer & Explanations
1. Evaluate 50! = 50*49*48* [47!] = 50*49*48 = 117600
47! 47!
2. 85P3 = 85! = 85! = 85*84*83*82! = 85*84*83 = 595650
[85-3]! 82! 82!
3. 20C20 = 1
[20C2]*[20C20] = 20! * 1= 20*19*18! = 20*19*1= 380
18! 18!
4. Exp: The word ‘ORANGE’ contains 6 different letters.
Therefore, Required number of words= Number of arrangement of 6 letters, taken all at a time
=6P6 = 6!= 6*5*4*3*2*1= 720
5. Exp: The total number of stations= 30
From 30 Stations we have to choose any two stations and the direction of travel [Ernakulam to Chennai
is different from Chennai to Ernakulam] in 30P2 ways.
30P2= 30*29= 870
6. Exp: The word ‘TECHNOLOGY’ contains 10 letters namely 2O, 1T, 1E, 1C, 1H, 1N, 1L, 1G, 1Y.
Therefore, Required number of ways= 10! = 10! [2!][1!] [1!] [1!] [1!] [1!] [1!] [1!] [1!] 2!
= 10*9* 8*7*6*5*4*3*2*1 = 1814400
2*1
7. Exp: Total number of balls= 2+3+5 = 10
2 balls can be drawn from 10 balls in 10C2
8. Exp: In the word ‘LANGUAGE’ we treat the vowels AUAE as one letter. Thus, we have LNGG [AUAE].
This we have 5 letters of which G occurs 2 times and the rest are different.
Number of ways arranging these letters= 5! =5*4*3= 60
स्रोत : placement.freshersworld.com
Answer: Total no. Of letters in Technology wword are 10 with O being repeated twice . So total no of arrangements =10!/2!=1,814,400.
Mathemagics & Puzzles
Mathematics, Vedic Math, Quantitative Aptitude Q&A, Puzzles.
In how many ways can the letters of the word, ‘TECHNOLOGY’ be arranged?
1 Answer Rao Sapna
Studied at Ramjas College, University of Delhi [Graduated 2022]2y
Total no. Of letters in Technology wword are 10 with O being repeated twice . So total no of arrangements =10!/2!=1,814,400.
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In how many different ways can the letters of the word 'CASUAL' be arranged?
In how many different ways can the letters of the word 'CASUAL' be arranged?
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In how many different ways can the letters of the word 'CASUAL' be arranged?
Question
In how many different ways can the letters of the word 'CASUAL' be arranged?
A 36 B 720 C 240 D 360 E None of these Open in App Solution
The correct option is E 360
The word CASUAL has 6 letters in which letter 'A' domes twice,
∴
Number of arrangements
= 6 ! 2 ! = 6 × 5 × 4 × 3 × 2 × 1 1 × 2 = 360 Suggest Corrections 0
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