<p>When performing a trick a magician says the word ABRACADABRA. The $11$ letters of this word are arranged in a row
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<ol class="abc-list"><li><div class="list-item-paragraph">Find the number of different arrangements that can be made.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Find the number of different arrangements in which the $2$ B’s are next to each other, the $2$ R’s are next to each other, exactly $4$ of the A’s are next to each other, and the C is next to the D.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">Given that the $11$ letters are arranged randomly, find the probability that all $5$ A’s are together.</div><p style="text-align: right;">[3]</p></li></ol>

When performing a trick a magician says the word ABRACADABRA. The $11$ letters of this word are arranged in a row


Find the number of different arrangements that can be made.

Find the number of different arrangements in which the $2$ B’s are next to each other, the $2$ R’s are next to each other, exactly $4$ of the A’s are next to each other, and the C is next to the D.

Given that the $11$ letters are arranged randomly, find the probability that all $5$ A’s are together.

Permutation and Combination

H2 Math

International Baccalaureate Higher Level Mathematics

IB HL Math

H1 Math

N2021 P2 Q7

5 A's, 2 B's, 2 R's, 1 C, 1 D = 11 letters

No. of ways $= \dfrac{11!}{5!2!2!} = 83160$

Arrangement: [B][B][R][R][D/C] with AAAA and A as separate blocks

No. of ways $= 3! \times {}^4C_2 \times 2! \times 2! = 144$

Arrangement: [B][R][B][D][A][A][A][A][R][C] treated as 7 distinct items

Required Prob $= \dfrac{\frac{7!}{2!2!}}{83160} = \dfrac{1260}{83160} = \dfrac{1}{66}$

N2021 P2 Q7

N2021 P2 Q7

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