(B) There are $6$ objects in total. We are given that $O_1$ and $O_2$ must occupy the $2$ bottom-most positions.These $2$ objects can be arranged amon

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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.These $2$ objects can be arranged among themselves in $2!$ ways.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 ways is $2! \times 4!$.

Competitive Quantitative Aptitude Permutation and Combination Permutation and Combination

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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Permutation and Combination

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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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The solution set of $^{10}C_{x-1} > 2 \cdot ^{10}C_x$ is

In how many ways can $6$ apples be distributed among $3$ boys,there being no restriction to the number of apples each boy may get?

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