How many numbers of 6 digits can be formed from the digits 1, 1, 2, 2, 3, 3?
(a) 30
(b) 60
(c) 90
(d) 120
Answer
627.9k+ views
Hint: To find the numbers that can be formed using 6 digits which are repetitive, we need to divide the permutation by factorial of the number of repetitive terms.
Complete step-by-step answer:
So, here we are given 6 digits 1, 1, 2, 2, 3, 3. We have to find out how many 6 digits numbers can be made from these digits. We know to find the total number of numbers possible using 6 digits we have to use a permutation combination.
In a 6 digit number we have spots for the 6 digits
_ _ _ _ _ _
As we have 6 digits with us (1, 1, 2, 2, 3, 3) so to fill the first position we have 6 options.
6 _ _ _ _ _
Next we are left with 5 digits, so to fill the second position we have 5 options with us.
6 5 _ _ _ _
Similarly, next we will be left with 4 then 3 then 2 and 1 options.
654321
Therefore, the total number of numbers possible $=6\times 5\times 4\times 3\times 2\times 1=720$
But 1, 2, 3 are repetitive in nature and they are repeated 2 times each.
We know to find the permutation of repetitive numbers we need to divide the whole permutation by factorial of the number of repetitive digits.
Therefore, total permutations
\[=\dfrac{6\times 5\times 4\times 3\times 2\times 1}{2!\times 2!\times 2!}\] (Because 1, 2, 3 are repeated 2 times each)
\[=\dfrac{6\times 5\times 4\times 3\times 2\times 1}{2\times 2\times 2}\] (Because $2!=2\times 1=2$)
= 90
Total number of numbers formed from 1, 1, 2, 2, 3, and 3 is 90.
Option (c) is the correct answer.
Note: A permutation is an arrangement of all or part of a set of objects, with regard to the order of the arrangement. Generally, people make a mistake by forgetting to divide the whole permutation by factorial of the number of repeated terms. This thing must be kept in mind to prevent losing unnecessary marks.
Complete step-by-step answer:
So, here we are given 6 digits 1, 1, 2, 2, 3, 3. We have to find out how many 6 digits numbers can be made from these digits. We know to find the total number of numbers possible using 6 digits we have to use a permutation combination.
In a 6 digit number we have spots for the 6 digits
_ _ _ _ _ _
As we have 6 digits with us (1, 1, 2, 2, 3, 3) so to fill the first position we have 6 options.
6 _ _ _ _ _
Next we are left with 5 digits, so to fill the second position we have 5 options with us.
6 5 _ _ _ _
Similarly, next we will be left with 4 then 3 then 2 and 1 options.
654321
Therefore, the total number of numbers possible $=6\times 5\times 4\times 3\times 2\times 1=720$
But 1, 2, 3 are repetitive in nature and they are repeated 2 times each.
We know to find the permutation of repetitive numbers we need to divide the whole permutation by factorial of the number of repetitive digits.
Therefore, total permutations
\[=\dfrac{6\times 5\times 4\times 3\times 2\times 1}{2!\times 2!\times 2!}\] (Because 1, 2, 3 are repeated 2 times each)
\[=\dfrac{6\times 5\times 4\times 3\times 2\times 1}{2\times 2\times 2}\] (Because $2!=2\times 1=2$)
= 90
Total number of numbers formed from 1, 1, 2, 2, 3, and 3 is 90.
Option (c) is the correct answer.
Note: A permutation is an arrangement of all or part of a set of objects, with regard to the order of the arrangement. Generally, people make a mistake by forgetting to divide the whole permutation by factorial of the number of repeated terms. This thing must be kept in mind to prevent losing unnecessary marks.
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