(B) There are $6$ objects in total. We are given that $O_1$ and $O_2$ must occupy the $2$ bottom-most positions. The number of ways to arrange $O_1$ a

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(B) There are $6$ objects in total. We are given that $O_1$ and $O_2$ must occupy the $2$ bottom-most positions. The number of ways to arrange $O_1$ and $O_2$ in these $2$ positions is $2!$. The remaining $4$ objects $(O_3, O_4, O_5, O_6)$ can be arranged in the remaining $4$ positions in $4!$ ways. Therefore,the total number of arrangements is $2! \times 4!$.

Class 11 Mathematics Permutation and Combination Definition of permutation, Number of permutations with or without repetition, Conditional permutations

Six objects $O_1$ to $O_6$ are arranged one on top of the other. In how many ways can these be arranged such that $O_1$ and $O_2$ are the $2$ bottom-most objects?

A
$4!$
B
$4! \times 2!$
C
$\frac{6!}{2!}$
D
$6!$

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Class 11

Mathematics Questions

Chapter

Permutation and Combination

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Definition of permutation, Number of permutations with or without repetition, Conditional permutations

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MediumWBJEE 2016

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Six objects $O_1$ to $O_6$ are arranged one on top of the other. In how many ways can these be arranged such that $O_1$ and $O_2$ are the $2$ bottom-most objects?

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In how many ways can the letters of the word $EAMCET$ be arranged such that no two vowels are together?

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