How many 3-digit numbers can be formed from 3 6 8 and 9 if repetition is not allowed
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(i) The possible 3 digit numbers using the digits 6, 0, 4 when repetition of digits is not allowed are 604, 640, 406 and 460. Show (ii) The possible 3 digit numbers using the digits 6, 0, 4 when repetition of digits is allowed are 400, 406, 460, 466, 444, 404, 440, 446, 464, 600, 604, 640, 644, 646, 664, 606, 660 and 666. How many 3 - digit numbers can be formed by using the digits 1 to 9 if repetition of digits is not allowed?
Answer (Detailed Solution Below)Option 1 : 504 Free 10 Questions 10 Marks 6 Mins Concept: Permutation: Permutation is defined as an arrangement of r things that can be done out of total n things. This is denoted by \({\;^n}{P_r} = \frac{{n!}}{{\left( {n - r} \right)!}}\) Calculation: Here order matters for example 123 and 132 are two different numbers. Therefore, there will be as many 3 digit numbers as there are permutations of 9 different digits taken 3 at a time. Therefore, the required 3 digit numbers ⇒ In 3rd place required number = 9 ⇒ in 2nd place required number = 8 ⇒ In 1st place required number = 7 So, 3 - digit number form by ⇒ 9 × 8 × 7 ⇒ 504
Combination: The number of selections of r objects from the given n objects is denoted by nCr = \(\rm \frac{n!}{r! (n - r)!}\) Last updated on Oct 19, 2022 The Application Links for the DSSSB TGT will remain open from 19th October 2022 to 18th November 2022. Candidates should apply between these dates. The Delhi Subordinate Services Selection Board (DSSSB) released DSSSB TGT notification for Computer Science subject for which a total number of 106 vacancies have been released. The candidates can apply from 19th October 2022 to 18th November 2022. Before applying for the recruitment, go through the details of DSSSB TGT Eligibility Criteria and make sure that you are meeting the eligibility. Earlier, the board has released 354 vacancies for the Special Education Teacher post. The selection of the DSSSB TGT is based on the Written Test which will be held for 200 marks. Answer: 64 ways Step-by-step explanation: If repetition is allowed:
Given that the numbers are 3,6,8, and 9 thus , n = 4 , taken at r = 3 digits P = n^r P = 4^3 P = 64 How many 3There are 504 different 3-digit numbers which can be formed from numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 if no repetition is allowed. Note: We can also use the multiplication principle to answer this question.
How many 3Therefore, '6' three-digit numbers can be formed.
Hence, the correct option is (d). How many combinations of 3 numbers can you have with repetition?There are 6 = 3x2x1 ways to order 3 digits in a row. Thus the number of combinations of 3 of the 10 digits is 720/6 = 120 combinations.
How many 3So, without repeating, six 3- digit number are formed. Total 6 + 6 = 12. Hence 12, 2-digit & 3-digit numbers can be formed by using the digits 3, 5, 6 without repeating any digit. KPSC Group C exam dates announced!
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