[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.
[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?
- 504
- 500
- 3024
- 336
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 = \[\rm \frac{n!}{r! [n - r]!}\]
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Answer:
64 ways
Step-by-step explanation:
If repetition is allowed:
- P = n^r
- Let n be the number of objects , taken at r at a time
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 3
How many 3
Hence, the correct option is [d].