
Seven digits from the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 are written down non-repeatedly in a random order to form a seven digit number. The probability that this seven digit number is divisible by 9 is
Answer
597.9k+ views
Hint: One major formula that might be used in the question is that
The method of choosing r number of different objects from n different objects is by using the formula
\[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
Another important formula is that
For arranging n different objects, the formula that can be used to find number of ways to do so is
Number of ways
\[=n!\]
Another aspect of the above formula is that if there are ‘r’ same objects in the ‘n’ different ones, then the formula become
\[=\dfrac{n!}{r!}\]
Complete step-by-step answer:
Now, in the question it is mentioned, we have to form numbers of 6 digits using the 4 digits such that the digits might get repeated.
There will be 2 cases that will form which are as follows
Case I. 3 different, 3 alike:- Only one number out of four will appear three times. Now using the formula given in the hint, we can write
\[{}^{4}{{C}_{1}}\]=4 ways.
Now, using the formula given in the hint, we have a set of 6 digits out of which three are alike and they can be arranged as
\[\begin{align}
& =\dfrac{6!}{3!} \\
& =120~ways \\
\end{align}\]
Therefore, by fundamental theorem, the number of numbers that are of this form
\[\begin{align}
& =4\times 120 \\
& =480~numbers \\
\end{align}\]
Case II:- 2 alike, 2 alike, 2 different
From the 4 digits, we can select 2 pairs of alike using the formula given in the hint as
\[\begin{align}
& {}^{4}{{C}_{2}}=\dfrac{4!\cdot 3!}{2!} \\
& =6~ways \\
\end{align}\] Now we have a set of 6 digits out of which 2 are alike of one kind and 2 of other kinds.
They can be arranged using the formula given in the hint
\[\begin{align}
& =\dfrac{6!}{2!2!} \\
& =180~ways \\
\end{align}\]
Hence, by fundamental theorem, the number of numbers which are of this form
\[\begin{align}
& =6\times 180 \\
& =1800~numbers \\
\end{align}\]
Therefore, the total numbers are
\[\begin{align}
& ~=480+1080 \\
& =1560 \\
\end{align}\]
Note: The students can make an error if they don’t know about the fundamental theorem and the formula that are mentioned in the hint as without knowing this property, the student won’t be able to get to the correct answer.
The method of choosing r number of different objects from n different objects is by using the formula
\[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
Another important formula is that
For arranging n different objects, the formula that can be used to find number of ways to do so is
Number of ways
\[=n!\]
Another aspect of the above formula is that if there are ‘r’ same objects in the ‘n’ different ones, then the formula become
\[=\dfrac{n!}{r!}\]
Complete step-by-step answer:
Now, in the question it is mentioned, we have to form numbers of 6 digits using the 4 digits such that the digits might get repeated.
There will be 2 cases that will form which are as follows
Case I. 3 different, 3 alike:- Only one number out of four will appear three times. Now using the formula given in the hint, we can write
\[{}^{4}{{C}_{1}}\]=4 ways.
Now, using the formula given in the hint, we have a set of 6 digits out of which three are alike and they can be arranged as
\[\begin{align}
& =\dfrac{6!}{3!} \\
& =120~ways \\
\end{align}\]
Therefore, by fundamental theorem, the number of numbers that are of this form
\[\begin{align}
& =4\times 120 \\
& =480~numbers \\
\end{align}\]
Case II:- 2 alike, 2 alike, 2 different
From the 4 digits, we can select 2 pairs of alike using the formula given in the hint as
\[\begin{align}
& {}^{4}{{C}_{2}}=\dfrac{4!\cdot 3!}{2!} \\
& =6~ways \\
\end{align}\] Now we have a set of 6 digits out of which 2 are alike of one kind and 2 of other kinds.
They can be arranged using the formula given in the hint
\[\begin{align}
& =\dfrac{6!}{2!2!} \\
& =180~ways \\
\end{align}\]
Hence, by fundamental theorem, the number of numbers which are of this form
\[\begin{align}
& =6\times 180 \\
& =1800~numbers \\
\end{align}\]
Therefore, the total numbers are
\[\begin{align}
& ~=480+1080 \\
& =1560 \\
\end{align}\]
Note: The students can make an error if they don’t know about the fundamental theorem and the formula that are mentioned in the hint as without knowing this property, the student won’t be able to get to the correct answer.
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