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- Question 6(vi-ix), Exercise 6.1
- $ (n!+1)$ is not divisible by any natural number between $2$ and $n$. ** Solution. ** We know $$n!=n(n-1... s 3.2.1$$ Hence $n!$ is divisible by every number between $1$ and $n$.\\ $n!$ can also divides by any natural number between $2$ and $n$.\\ For $(n!+1)$, $1$ is not divisible by any natural number between $2$ and $n$.\\ So $ (n!+1)$ is not divisible by a
- Exercise 6.1 (Solutions)
- ) $(n!+1)$ is not divisible by any natural number between 2 and $n$. (ix) $\quad(n!)^{2} \leq n^{n} . n!<(2
- Question 8 and 9, Exercise 6.2
- -1)! arrangements\\ while $4$ men can be adjusted between seats of women in $41$ ways,\\ so total possible
- Question 20 and 21, Exercise 6.2
- ssible arrangements\\ and $6$ men may be adjusted between in $6!$ possible arrangements\\ and hence total p
- Question 9 and 10, Exercise 6.3
- etermined by $n$-points \\ and all possible lines between $n$ points is $C_{2}$\\ but $n$ lines are sides o