
The number of ways in which 10 candidates $A1, A2, A3, A4......A10$ can be ranked if $A1$ is just above $A2$, then the number of ways is
A. 9!2!
B. 10!
C. 10!2!
D. 9!
Answer
439.1k+ views
Hint: We are given that there are total 10 candidates. We have to find the number of ways in which 10 candidates can be ranked if $A1$ is just above $A2$. Therefore, fix $A1$ and $A2$ as a single unit and now arrange the candidates for 9 ranks. The number of ways of arranging $n$ distinct objects in $n$ places is $n!$
Complete step by step Answer:
There is a total of 10 candidates who are to be ranked.
We want $A1$to come just above $A2$.
Let us take $A1$ and $A2$ as a single entity. Then, both of these will always come together.
Also, there is only one possible arrangement for $A1$ and $A2$.
Next, now we have to arrange 9 entities in 9 places.
The first rank can be given to any of the 9 candidates and similarly, the second rank can be given to 8 candidates and so on.
The number of ways of arranging $n$ distinct objects in $n$ places is $n!$
Since no candidate can take the same position twice, then the number of ways in the 9 entities can be arranged is 9!
Therefore, the number of ways in which 10 candidates $A1, A2, A3, A4......A10$ can be ranked if $A1$ is just above $A2$ is 9!
Hence, option D is correct.
Note: Some students can mistake by not fixing $A1$ and $A2$ together. There are methods such as combination and permutation which helps to calculate the possible number of ways. When the order of the arrangements matters, we use permutation and when the order does not matter, then the combination is used.
Complete step by step Answer:
There is a total of 10 candidates who are to be ranked.
We want $A1$to come just above $A2$.
Let us take $A1$ and $A2$ as a single entity. Then, both of these will always come together.
Also, there is only one possible arrangement for $A1$ and $A2$.
Next, now we have to arrange 9 entities in 9 places.
The first rank can be given to any of the 9 candidates and similarly, the second rank can be given to 8 candidates and so on.
The number of ways of arranging $n$ distinct objects in $n$ places is $n!$
Since no candidate can take the same position twice, then the number of ways in the 9 entities can be arranged is 9!
Therefore, the number of ways in which 10 candidates $A1, A2, A3, A4......A10$ can be ranked if $A1$ is just above $A2$ is 9!
Hence, option D is correct.
Note: Some students can mistake by not fixing $A1$ and $A2$ together. There are methods such as combination and permutation which helps to calculate the possible number of ways. When the order of the arrangements matters, we use permutation and when the order does not matter, then the combination is used.
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