Find the number of different ways in which the 10 letters of the word SUMMERTIME can be arranged so that the Es are not together. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mathematics
IGCSE&ALevel
CAIE
Exam No:9709_s20_qp_52 Year:2020 Question No:6(b)

Answer:

Total number $=\frac{10 !}{2 ! 3 !}(302400) \quad(\mathrm{A})$
With Es together $=\frac{9 !}{3 !}(60480) \quad(B)$
Es not together $=$ their $(\mathrm{A})-$ their $(\mathrm{B})$
241920
Alternative method for question $6(b)$
${ }{ }_{-}{ }^{\wedge}{ }_{-}{ }^{\wedge}{ }_{-}{ }^{\wedge}{ }_{-}{ }^{\wedge}{ }_{-}{ }^{\wedge}{ }_{-}{ }^{\wedge}{ }_{-}{ }^{\wedge}{ }_{-}{ }^{\wedge}{ }_{-}$
$
\frac{8 !}{3 !} \times \frac{9 \times 8}{2}
$
$8 ! \times k$ in numerator, $k$ integer $\geqslant 1$, denominator $\geqslant 1$
$3 ! \times m$ in denominator, $m$ integer $\geqslant 1$
Their $\frac{8 !}{3 !}$ Multiplied by ${ }^{9} \mathrm{C}_{2}$ (OE) only (no additional terms)
241920

Knowledge points:

5.2.2.1 repetition (e.g. the number of ways of arranging the letters of the word ‘NEEDLESS’)
5.2.2.2 restriction (e.g. the number of ways several people can stand in a line if two particular people must, or must not, stand next to each other). (Questions may include cases such as people sitting in two (or more) rows.) (Questions about objects arranged in a circle will not be included.)

Solution:

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