Question 9 & 10 Review Exercise 6

Solutions of Question 9 & 10 of Review Exercise 6 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

How many numbers greater than a million can be formed with the digits $2,3,0,3,4,2,3$?

We know that $1$ million $=100,0000$.

First we are computing the total number of ways arranging these digits using repeated permutation as: $$=\dfrac{7 !}{3 ! \cdot 2 !}=420$$ But we have find the total number that are greater than $1$ million.

In this case number should not start with $0$,

therefore the total numbers that do not start with zero.

It can be formed using the given $7$ digits taking out $0$ from them are: $$=\dfrac{6 !}{2 ! 3 !}=60$$ Thus the total numbers greater than $1$ million are $420-50=360$.

A party of $n$ men is to be seated around a circular table. Find the probability that two particular men sit together.

The total number of ways that $n$ persons can be seated around a circular table are: $(n-1)$ !

If two persons sit together, we shall deal these two as one man, then in this case total will be $(n - 1)$.

and the total number of ways these $(n-1)$ can sit around table are: $(n-2) !$

the number of ways that $2$ men can sit are: $2 !$

Thus by fundamental principle of counting the total number of ways of sitting these $n$ people in which two men want to sit together are: $$(n-2) ! \cdot 2 !=2(n-2) !$$

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