(B) There are $7$ scientists in total. The total number of ways to arrange $7$ lectures is $7!$. In any arrangement,the relative order of $S_1, S_3,$

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(B) There are $7$ scientists in total. The total number of ways to arrange $7$ lectures is $7!$. In any arrangement,the relative order of $S_1, S_3,$ and $S_7$ can occur in $3! = 6$ possible ways. These $6$ ways are: $(S_1, S_3, S_7), (S_1, S_7, S_3), (S_3, S_1, S_7), (S_3, S_7, S_1), (S_7, S_1, S_3), (S_7, S_3, S_1)$. Out of these $6$ ways,only one way satisfies the condition that $S_1$ is prior to $S_3$ and $S_3$ is prior to $S_7$ (i.e.,the order $S_1 Therefore,the required number of ways is $\frac{7!}{3!} = \frac{5040}{6} = 840$.

Class 11 Mathematics Permutations and Combinations Definition of permutation, Number of permutations with or without repetition, Conditional permutations

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Seven scientists $S_1, S_2, \ldots, S_7$ are invited to deliver one lecture each in a conference. The number of ways all the seven lectures can be arranged such that the lecture of $S_1$ is prior to that of $S_3$ and the lecture of $S_3$ is prior to that of $S_7$ is

TS EAMCET 2021 All TS EAMCET

A
$35$
B
$840$
C
$720$
D
$210$

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

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Seven scientists $S_1, S_2, \ldots, S_7$ are invited to deliver one lecture each in a conference. The number of ways all the seven lectures can be arranged such that the lecture of $S_1$ is prior to that of $S_3$ and the lecture of $S_3$ is prior to that of $S_7$ is

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