(CIE/0606/2021/m/22Q8) A photographer takes 12 different photographs. There are 3 of sunsets, 4 of oceans, and 5 of mountains.

(a) The photographs are arranged in a line on a wall.

(i) How many possible arrangements are there if there are no restrictions?$[1]$

(ii) How many possible arrangements are there if the first photograph is of a sunset and the last photograph is of an ocean? $\quad[2]$

(iii) How many possible arrangements are there if all the photographs of mountains are next to each other?

(b) Three of the photographs are to be selected for a competition.

(i) Find the number of different possible selections if no photograph of a sunset is chosen. $\quad[2]$

(ii) Find the number of different possible selections if one photograph of each type (sunset, ocean, mountain) is chosen.

*****\\

*********math solution*************

$\begin{array}{lll}\text{(a)}&\text{(i)}&\text{There are total of 12 photographs.}\\&& \text{Number of arrangements }=12 !=479001600 \\&\text {(ii)}&\text{First place (3 sunset photographs)}= 3 \text { ways } \\&&\text {Last place (4 ocean photographs) }=4 \text { ways } \\&&\text {Total number of arrangements }= 3 \times 10 ! \times 4=43545600 \\&\text{(iii) }&\text {Sunset and ocean photographs }=7 \\&&\text {Number of arrangements }=7 ! \\&&\text {Number of arrangements for mountains }=5 ! \\&&\text {There are } 6 \text { places of between } 7 \text { photographs } \\&&\text {Thus total of } 8 \text { places including left and right places } \\&&\text {Therefore total number of arrangements }=8 \times 7 ! \times 5 ! \\&&=4838400\end{array}$

(b) $\begin{array}[t]{rl}\text{(i)}&\text{Number of ocean and mountain photographs}=9\\&\text{Thus total number of selections }=^9C_3=84\\\text{(ii)}&\text{Choose 1 from each type}\\&\text{Total number of selections }=^3C_1\times ^4C_1\times ^5C_1=60\end{array}$
**********end math solution********************

## 0 Comments