How many 3-digit numbers can be formed from the digits 1, 2, and 3 if repetitions are allowed?
So basically, I attempted this question as- There are 4 numbers and 3 places to put in the numbers: In the ones place, any 4 numbers can be put, so there are 4 choices in the ones place. Similarly for the tens and the hundreds place. So, the total choices are, by multiplication principle- $$4*4*4=64$$ And well and good, this was the answer. But what if I reversed the method? So I take some particular numbers, like $1,2,3$ and say that, well, $1$ can go in $3$ places, $2$ in $2$ places and $3$ in $1$ place, so by multiplication principle, there are $6$ ways of forming a $3$-digit number with $1,2,3$. But there are $4$ different numbers. So the number of $3$-number combinations are- $(1,2,3)$,$(1,2,4)$,$(1,3,4)$,$(2,3,4)$. Each can be arranged in $6$ ways, so we get $24$ ways totally. So why is my answer different here? How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that:1) Repetition is not allowed.2) Repetition is allowed.Answer Verified Hint: Fundamental principle of counting: According to the fundamental principle of counting if a task can be done in “m” ways and another task can be done in “n” ways, then the number of ways in which both the tasks can be done in mn ways. Complete step by step solution: Note: The number of 3-digit numbers formed using the digits 1, 2, 3, 4 and 5 when repetition is not allowed is equivalent to the number of 3 Letter permutations of 5 distinct letter = ${}^{5}{{P}_{3}}=\dfrac{5!}{\left( 5-3 \right)!}=\dfrac{5!}{2!}=\dfrac{120}{2}=60$ which is the same result as above. How many 3So I take some particular numbers, like 1,2,3 and say that, well, 1 can go in 3 places, 2 in 2 places and 3 in 1 place, so by multiplication principle, there are 6 ways of forming a 3-digit number with 1,2,3. But there are 4 different numbers. So the number of 3-number combinations are- (1,2,3),(1,2,4),(1,3,4),(2,3,4).
How many 3Problem 2: How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated? Solution: Answer: 108.
How many 3Therefore, 0 three digit prime numbers can be formed using 1,2 and 3 without repetition.
How many different numbers of three digits can be formed with the digits 1 2 3 4 5 no digit is being repeated?The answer is 1024. 11111 11112 11113 11114 11121 11122 11123 11124 11131 11132 11133 11134 11141 11142 11143 11144 11211 11212 11213 11214 11221 11222 11223 11224 11231 11232 11233 11234 11241 11242 …. How many even, 5 digit numbers can be formed from digits 3, 4, 5, 6, and 7?
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