InterviewSolution
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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Find the number of integers greater than 7000 that can be formed with the digits 3, 5, 7, 8 and 9 where no digits are repeated. |
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Answer» We have to find the number of integers greater than 7000 with the digits 3,5, 7, 8 and 9. So, with these digits, we can make maximum five-digit numbers because repetition is not allowed. Since all the five-digit numbers are greater than 7000, we have Number of five-digit integers = 5 x 4 x 3 x 2 x 1 = 120 A four-digit integer is greater than 7000 if thousandth place has any one of 7, 8 and 9. Thus, thousandth place can be filled in 3 ways. The remaining three places can be filled from remaining four digits in 4P3 ways. So, total number of four-digit integers = 3x 4P3 = 3 x 4 x 3 x 2 = 72 Total number of integers = 120 + 72 = 192 |
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| 2. |
Find the number of positive integers greater than 6000 and less than 7000 which are divisible by 5, provided that no digit is to be repeated. |
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Answer» Thousand Place is to be fill with 6. Number of way = 1 ten place can be fill with 7 digits. Number of ways= 7. Thus, required number will be =1 x 8 x 7 x 2 =112 |
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| 3. |
Prove that 33! Is divisible by 215, what is the largest integer nsuch that 33! Is divisible by 2n? |
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Answer» 33! = 33 × 32 × 31 × 30 × 29 … × 16 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 33 × (2)5× 31 × 30 × 29 × … × (2)4… × (2)3× 7 × 6 × 5 × (2)2× 3 × (2)1× 1 Now take all 2’s out we get 33! = (2)5∙ (2)4∙ (2)3 ∙ (2)2∙ (2)1[33 × 31 × 30 × 29 × … × 1] [33 × 31 × 30 × 29 × … × 1] contains numbers 6, 10, 12, 14, 18, 20, 22, 24, 26, 28, 30, which have 2 as one of their factors. So, the product of 2 present in these numbers = 2 × 2 × 22× 2 × 2 × 22× 2 × 23× 2 × 22× 2 = 216 Therefore, we have 215× 216= 231 Thus, 31 is the largest integer such that 33! Is divisible by 2n. |
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| 4. |
In how many ways can the letters of the word ‘PENCIL’ be arranged so that I is always next to L. |
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Answer» There are 6 letters in the word ‘PENCIL’ Consider LI as one letter. Now 5 Letters (P,E,N,C,I,L) can be arranged in\(^5P_5=5! =120\) ways. Hence, the total number of ways in which I is always next to L is 120. |
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| 5. |
How many five-digit numbers formed using the digit 0, 1, 2, 3, 4, 5 are divisible by 5 if digits are not repeated? |
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Answer» Here, repetition of digits is not allowed. For a number to be divisible by 5, unit’s place digit should be 0 or 5. Case I: when unit’s place is 0 Unit’s place digit can be selected in 1 way. 10’s place digit can be selected in 5 ways. 100’s place digit can be selected in 4 ways. 1000’s place digit can be selected in 3 ways. 10000’s place digit can be selected in 2 ways. ∴ Total number of numbers = 1 × 5 × 4 × 3 × 2 = 120. Case II: when the unit’s place is 5 Unit’s place digit can be selected in 1 way. 10000’s place should be a non-zero number. ∴ It can be selected in 4 ways. 1000’s place digit can be selected in 4 ways. 100’s place digit can be selected in 3 ways. 10’s place digit can be selected in 2 ways. ∴ Total number of numbers = 1 × 4 × 4 × 3 × 2 = 96 ∴ Total number of required numbers = 120 + 96 = 216 |
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