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1). In how many different ways, the letters of the word 'INHALE' can be arranged?
View Answer Correct Answer: 720 The word 'INHALE' has 6 distinct letters. Number of arrangements = n! = 6! = 6 X 5 X 4 X 3 X 2 X 1 = 720
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2). In how many different ways, the letters of the word 'ARMOUR' can be arranged?
View Answer Correct Answer: None of the above Number of arrangements = \( \large\frac{n!}{p! q! r!} \) Total letters = 6, but R has come twice So, required number of arrangements = \( \large\frac{6!}{2!} = \frac{6 \times 5 \times 4 \times 3 \times 2!}{2!} = 360\)
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3). In how many different ways, the letters of the word 'BANKING' can be arranged?
View Answer Correct Answer: 2520 Total letters = 7, but N has come twice. So, the required number of arrangements = \( \large\frac{7!}{2!} = \frac{7 \times 6 \times 5 \times 4 \times 3 \times 2!}{2!} = 2520\)
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4). In how many different ways, can the letters of the word 'VENTURE' be arranged?
View Answer Correct Answer: 2520 The required different ways = \( \large\frac{7!}{2!} = \frac{7 \times 6 \times 5 \times 4 \times 3 \times 2!}{2!} = 2520\) | ||||||||

5). How many different signals , can be made by 5 flags from 8 flags of different colors?
View Answer Correct Answer: 6720 The number of ways taking 5 flags out of 8 flags. =\( 8P_5 = \frac{8!}{(8 - 5)!} = \frac{8!}{3!}\) = \( \large \frac{8 \times 7 \times 6 \times 5 \times 4 \times 3!}{3!} = 6720\) | ||||||||

6). A child has four pockets and three marbles. In how many ways, the child can put the marbles in the pockets?
View Answer Correct Answer: 64 The first marble can be put into the pockets in 4 ways, so can the second and third. Thus, the number of ways in which the child can put the marbles = 4 X 4 X 4 = 64 ways | ||||||||

7). In how many different ways, can the letters of the word 'ASSASSINATION' be arranged, so that all S are together?
View Answer Correct Answer: 151200 When all the S are taken together, then AS^AS^INATION are letters. So, 10 letters in the total can be arranged in 10 ways. All S are considered as 1 But here are 3 'A' and 2 'I' and 2 'N'. The required number of ways = \( \large\frac{10!}{3! \times 2! \times 2!} = 151200 \) | ||||||||

8). There is a 7-digit telephone number with all different digits. If the digit at extreme right and extreme left are 5 and 6 respectively, find how many such telephone numbers are possible?
View Answer Correct Answer: 6720 There is a 7-digit telephone number but extreme right and extreme left positions are fixed ie.., 6 x x x x x 5 So the required number of ways = 8 x 7 x 6 x 5 x 4 = 6720 | ||||||||

9). In a meeting between two countries, each country has 12 delegates, all the delegates of one country shakes hands with all delegates of the other country. Find the number of handshakes possible?
View Answer Correct Answer: 144 Total number of handshakes = 12 x 12 = 144 | ||||||||

10). Find the number of ways, in which 12 different beads can be arranged to form a necklace.
View Answer Correct Answer: 11! / 2 Number of arrangements of beads = (12 -1)! = 11!, but it is not mentioned that either it is clockwise or anti-clockwise. So, required number of arrangements = \( \large\frac{1}{2}(12 - 1)! = \frac{11!}{2} \) |

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