Recent questions tagged combination

Description : In how many ways can 8 different balls be distributed in 6 different boxes can contain any number of balls except that ball 4 can only be put into box 4 or 5 ? A) 2×5^6 B) 2×6^7 C) 2×5^4 D) 2×4^7

Last Answer : Answer: B)  1st ball can be put in any of the 6 boxes.  2nd ball can be put in any of the 6 boxes.  3rd ball can be put in any of the 6 boxes.  Ball 4 can only be put into box 4 or box 5. Hence, 4th ball ... put in any of the 6 boxes.  Hence, required number of ways = 6 6 6 2 6 6 6 6  = 2 6^7

Description : A school has 9 maths teachers and 6 science teachers. In how many ways can a team of 4 maths teachers be formed from them such that the team must contain exactly 1 science teacher? A) 800 B) 720 C) 680 D) 504 

Last Answer : Answer: D) The team should have 4 maths teachers. But the team must contain exactly 1 science teacher. Hence, select 3 maths teachers from 9 maths teachers and select 1 science teachers from 6 science teachers.  Number of ways this can be ...  ={9 8 7}/{3 2 1}X6  = 504 / 6 6  = 84 6 =504 

Description :  In how many ways can 10 stones can be arranged to form a bangles? A) 267720 B) 284360 C) 125380 D) 181440

Last Answer : Answer: D) Number of arrangements possible = {1} / {2} X (10-1)!  = {1} / {2}X 9! = {1} / {2} X 9×8×7× 6×5×4×3×2×1 = {1 / 2 } ×362880 = 181440

Description : In how many ways can 6 girls be seated in a rectangular order? A) 60 B) 120 C) 5040 D) 720

Last Answer : Answer: B)  Number of arrangements possible = (6-1)!  = 5! = 5×4×3×2×1 = 120

Description :  In how many ways can a team of 6 persons be formed out of a total of 12 persons such that 3 particular persons should not be included in any team? A) 56 B) 112 C) 84 D) 128

Last Answer : Answer: C) Three particular persons should not be included in each team. i.e., we have to select remaining 6- 3= 3 persons from 12-3 = 9 persons.  Hence, required number of ways = 9C3  = {9×8 × 7} / {3 × 2 × 1} = 504 / 6 = 84

Description : There are 10 orange, 2 violet and 4 purple balls in a bag. All the 16 balls are drawn one by one and arranged in a row. Find out the number of different arrangements possible. A) 25230 B) 23420 C) 120120 D) 27720

Last Answer : Answer: C)  Number of different arrangements possible  = {16!} / {10! 2! 4!}  = {16×15×14×13×12×11×10×9×8×7×6×5×4×3×2 } /  {(10×9×8×7×6×5×4×3×2 ) (2) (4×3×2)}}  = {16×15×14×13×12×11} / {(2)(4×3×2)}  = {8×5×7×13×3×11}  = 120120

Description : A bowl contains 8 violet, 6 purple and 4 magenta balls. Three balls are drawn at random. Find out the number of ways of selecting the balls of different colours? A) 362 B) 2 48 C) 122 D) 192

Last Answer : Answer: D)  1 violet ball can be selected is 8C1 ways.  1 purple ball can be selected in 6C1 ways.  1 magenta ball can be selected in 4C1 ways.  Total number of ways = 8C1 × 6C1 × 4C1  = 8×6×4  = 192

Description : In how many ways can 8 different ballons be distributed among 7 different boxes when any box can have any number of ballons? A) 5^4-1 B) 5^4 C) 4^5-1 D) 7^8

Last Answer : Answer: D) Here n = 7, k = 8. Hence, required number of ways = n^k  =7^8 

Description : A box contains 2 pink balls, 3 brown balls and 4 blue balls. In how many ways can 3 balls be drawn from the box, if at least one brown ball is to be included in the draw? A) 32 B) 48 C) 64 D) 96 E) None

Last Answer : Answer: C) 

Description : In how many ways can a group of 10 men and 5 women be made out of a total of 12 men and 10 women? A) 16632 B) 15290 C) 25126 D) 34845 E) 38135

Last Answer : Answer: A)  Required number of ways = 12C10 x 10C5  = 66 × 252 = 16632

Description : How many 5-letter code words are possible using last 10 letter of the English alphabet , if no letter can be repeated ? a) 30240 b) 25440 c) 45640 d) 32940

Last Answer : Answer: A)  The number of 5 letter code words out of the last 10 letters of the English alphabets are = 10× 9× 8 × 7× 6  = 80 × 63× 6  = 30240 ways.

Description : There are 7 periods in each working day of a college. In how many ways can one organize 6 subjects such that each subject is allowed at least one period? A) 33200 B) 15120 C) 10800 D) 43600

Last Answer : Answer: B)  6 subjects can be arranged in periods in 7P6 ways.  Remaining 1 period can be arranged in 6P1 ways.  Two subjects are alike in each of the arrangement. So we need to divide by 2! to avoid over counting. ... of arrangements = (7P6 x 6P1)/2!  = 5040 6 / 2 = 30240 / 2 = 15120

Description :  In how many different ways can 6 apple and 6 orange form a circle such that the apple and the orange alternate? A) 82880 B) 86400 C) 71200 D) 63212

Last Answer : Answer: B)  6 apples can be arranged in (6-1)! Ways  Now there are 6 positions in which 6 orange can be placed.  This can be done in 6! ways. Required number of ways = (6-1)! × 6!  = 5! × 6!  = 120 × 720  = 86400

Description : Find out the number of ways in which 12 Bangles of different types can be worn in 2 hands? A) 1260 B) 2720 C) 1225 D) 4096

Last Answer : Answer: D)  The first bangle can be worn in any of the 2 hands (2 ways). Similarly each of the remaining 11 bangles also can be worn in 2 ways. Hence total number of ways=2×2×2×2×2×2×2×2×2×2×2×2  =2^12  =4096

Description : How many 3-letter words can be formed with or without meaning from the letters A , G , M , D , N , and J , which are ending with G and none of the letters should be repeated? a) 20 b) 18 c) 25 d) 27

Last Answer : Answer: A) Since each desired word is ending with G, the least place is occupied with G. So, there is only 1 way. The second place can now be filled by any of the remaining 5 letters (A , M , D , N , J ... letters. So, there are 4 ways to fill. Required number of words = (1 x 5 x 4) = 20.

Description : How many 3 letters words (with or without meaning) can be formed out of the letters of the word, "PLATINUM", if repetition of letters is not allowed? a) 742 b) 850 c) 990 d) 336 

Last Answer : Answer: D)  The word PLATINUM contains 8 different letters. Required number of words = number of arrangements of 8 letters taking 3 at a time.  = 8p3  = 8 x 7 x 6  = 56×6  = 336

Description : In how many different ways can the letters of the word 'DILUTE' be arranged such that the vowels may appear in the even places? a) 36 b) 720 c) 144 d) 24

Last Answer : Answer: A)  There are 3 consonants and 3 vowels in the word DILUTE.  Out of 6 places, 3 places odd and 3 places are even.  3 vowels can arranged in 3 even places in 3p3 ways = 3! = 6 ways.  And then 3 ... 3 places in 3p3 ways = 3! = 6 ways.  Hence, the required number of ways = 6 x 6 = 36.

Description :  In how many different ways can the letters of the word "POMADE" be arranged in such a way that the vowels occupy only the odd positions? a) 72 b) 144 c) 532 d) 36

Last Answer : Answer: D)  There are 6 different letters in the given word, out of which there are 3 vowels and 3 consonants.  Let us mark these positions as under:  [1] [2] [3] [4] [5] [6]  Now, 3 vowels can be ... of these arrangements = 3P3  = 3!  = 6 ways.  Therefore, total number of ways = 6 x 6 = 36.

Description : In how many different ways can the letters of the word "XANTHOUS" be arranged in such a way that the vowels occupy only the odd positions? a) 2880 b) 4320 c) 2140 d) 5420

Last Answer : Answer: A) There are 8 different letters in the given word "XANTHOUS", out of which there are 3 vowels and 5 consonants.  Let us mark these positions as under:  [1] [2] [3] [4] [5] [6] [7] ... in 5P5 ways = 5! Ways  = 120 ways.  Therefore, required number of ways = 24 x 120 = 2880 ways.

Description : In how many different ways can the letters of the word "ZYMOGEN" be arranged in such a way that the vowels always come together? a) 1440 b) 1720 c) 2360 d) 2240

Last Answer : Answer: A)  The arrangement is made in such a way that the vowels always come together. i.e., "ZYMGN(OE)". Considering vowels as one letter, 6 different letters can be arranged in 6! ways; i.e., 6! = 720 ... in 2! ways; i.e.,2! = 2 ways Therefore, required number of ways = 720 x 2 = 1440 ways.

Description : How many different possible permutations can be made from the word ‘WAGGISH’ such that the vowels are never together? A) 3605 B) 3120 C) 1800 D) 1240 E) 2140

Last Answer : Answer: C)  The word ‘WAGGISH’ contains 7 letters of which 1 letter occurs twice  = 7! / 2! = 2520 No. of permutations possible with vowels always together = 6! * 2! / 2!  = 1440 / 2 = 720 No. of permutations possible with vowels never together = 2520-720  = 1800.

Description : In how many different ways can the letters of the word 'VINTAGE' be arranged such that the vowels always come together? A) 720 B) 1440 C) 632 D) 364 E) 546

Last Answer : Answer: A)  It has 3 vowels (IAE) and these 3 vowels should always come together. Hence these 3 vowels can be grouped and considered as a single letter. That is, VNTG(IAE). Hence we can assume total ... ways to arrange these vowels among themselves 3! = 3 2 1=6 Total number of ways 120 6=720

Description : In how many different ways can the letters of the word 'SPORADIC' be arranged so that the vowels always come together? A) 120 2 B) 1720 C) 4320 D) 2160 E) 2400

Last Answer : Answer: C) The word 'SPORADIC' contains 8 different letters. When the vowels OAI are always together, they can be supposed to form one letter. Then, we have to arrange the letters SPRDC (OAI). Now, 6 ... be arranged among themselves in 3! = 6 ways. Required number of ways = (720 x 6) = 4320.

Description : In how many different ways can the letters of the word 'RAPINE' be arranged in such a way that the vowels occupy only the odd positions? A) 32 B) 48 C) 36 D) 60 E) 120

Last Answer : Answer: C)  There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.  Let us mark these positions as under:  (1) (2) (3) (4) (5) (6)  Now, 3 vowels can be placed at ... .  Number of ways of these arrangements = 3P3 = 3! = 6.  Total number of ways = (6 x 6) = 36

Description :  In how many different ways can the letters of the word 'POTENCY' be arranged in such a way that the vowels always come together? A) 1360 B) 2480 C) 3720 D) 5040 E) 1440

Last Answer : Answer: E)  The word 'POTENCY' has 7 different letters.  When the vowels EO are always together, they can be supposed to form one letter.  Then, we have to arrange the letters PTNCY (EO).  Now, 6 (5 ... be arranged among themselves in 2! = 2 ways.  Required number of ways = (720 x2)  = 1440.

Description :  In how many different ways can the letters of the word 'ABOMINABLES' be arranged so that the vowels always come together? A) 181045 B) 201440 C) 12880 D) 504020 E) 151200

Last Answer : Answer: E)  In the word 'ABOMINABLES', we treat the vowels AOIAE as one letter.  Thus, we have BMNBLS (AOIAE).  This has 7 (6 + 1) letters of which B occurs 2 times and the rest are different. Number of ways arranging these letters = 7! / 2!  = (7×6×5×4×3×2×1) / (2×1) = 2520

Description : In how many different ways can any 4 letters of the word 'ABOLISH' be arranged? a) 5040 b) 840 c) 24 d) 120

Last Answer : Answer: B) There are 7 different letters in the word 'ABOLISH'.  Therefore,  The number of arrangements of any 4 out of seven letters of the word = Number of all permutations of 7 letters, taken 4 at a time = ... we have  7p4 = 7 x 6 x 5 x 4 = 840.  Hence, the required number of ways is 840.

Description : In how many different ways can the letters of the word 'GRINDER' be arranged? A) 2520 B) 1280 C) 3605 D) 1807 E) 1900

Last Answer : Answer: A)  In these 7 letters, 'R' occurs 2 times, and rest of the letters are different.  Hence, number of ways to arrange these letters  = {7!} / {(2!) }  = {7×6×5×4×3×2×1} / {2×1}  = 2520.

Description : How many arrangements can be made out of the letters of the word 'BIGBOSS' ? A) 9240 B) 2772 C) 1260 D) 1820 E) 2800

Last Answer : Answer: C)  The word 'BIGBOSS' has 7 letters  In these 7 letters, B(2) , I(1), G(1) , O(1),S(2)  Hence, number of ways to arrange these letters  = {7!} / (2!)(1!)(1!)(2!)}  = 5040/4  = 1260

Description : How many words can be formed by using all letters of the word 'CABIN'? A) 720 B) 24 C) 120 D) 60 E) None

Last Answer : Answer: C)  The word 'CABIN' has 5 letters and all these 5 letters are different.  Total number of words that can be formed by using all these 5 letters  = 5P5  = 5!  = 5×4×3×2×1  = 120 

Description : In how many ways can the letters of the word 'NOMINATION' be arranged? A) 237672 B) 123144 C) 151200 D) 150720 E) None of these

Last Answer : Answer: C)  The word 'NOMINATION' contains 10 letters, namely  3N, 2O, 1M, 2I,1A, and 1T. Required number of ways = 10 ! / (3!)(2!)(1!)(2!)(1!)(1!)  = 151200

Description : Calculate the value of equivalent capacitance of the combination.

Last Answer : Calculate the value of equivalent capacitance of the combination

Description : A solar pond is a combination of :  (A) Solar energy storage and heat collection  (B) Solar energy collection and heat storage  (C) Solar energy collection and energy storage  (D) All the above

Last Answer : A solar pond is a combination of : Solar energy collection and heat storage