## Question 7(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $6! = 720$ | B1 | Evaluated |

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## Question 7(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Total no. of arrangements: $\frac{9!}{2!3!} = 30240$ | B1 | Accept unevaluated |
| No. with Ts together $= \frac{8!}{3!} = 6720$ | B1 | Accept unevaluated |
| With Ts not together: $30240 - 6720$ | M1 | Correct or $\frac{9!}{m} - \frac{8!}{n}$, $m,n$ integers $>1$, or *their* identified total $-$ *their* identified Ts together |
| $23520$ | A1 | CAO |
| **Alternative method:** | | |
| $\frac{7!}{3!} \times \frac{8\times7}{2}$ | B1 | $7! \times (k>0)$ in numerator, cannot be implied by ${}^7P_2$ etc. |
| | B1 | $3! \times (k>0)$ in denominator |
| | M1 | $\frac{their\ 7!}{their\ 3!} \times {}^8C_2$ or ${}^8P_2$ |
| $23520$ | A1 | CAO |

## Question 7(iii):

**Answer:** Number of arrangements $= \frac{7!}{3!}$

Probability $= \dfrac{\text{their } \frac{7!}{3!}}{\text{their } \frac{9!}{3!2!}} = \frac{840}{30240}$ | M1 | *their* identified number of arrangements with T at ends ÷ *their* identified total number of arrangements; $\frac{7!}{3!}$ or $\frac{m}{9!}m, n$ integers $> 1$

$\dfrac{1}{36}$ or $0.0278$ | A1 | Final answer

**Total: 2 marks**

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## Question 7(iv):

$OOT_{--}$ $\quad {}^4C_2 = 6$ | M1 | ${}^4C_x$ seen alone or ${}^4C_x \times k \geq 1$, $k$ an integer, $0 < x < 4$

$OOTT_{-}$ $\quad {}^4C_1 = 4$

$OOOT_{-}$ $\quad {}^4C_1 = 4$

$OOOTT$ $\quad = 1$ | A1 | ${}^4C_2 \times k$, $k = 1$ or ${}^4C_1 \times m$, $m = 1$ oe alone

| M1 | Add 3 or 4 identified correct scenarios only, accept unsimplified

$(Total) = 15$ | A1 | CAO, WWW. Only dependent on 2nd M mark

**Total: 4 marks**