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A. $\dfrac{4}{{165}}$

B.$\dfrac{{16}}{{165}}$

C.$\dfrac{{12}}{{165}}$

D.$\dfrac{5}{{165}}$

Answer

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Hint: We 1st take count of all 11 letters and rearrange them in total number of outcomes and then take S’s as a single letter and then find the favorable outcomes.

Complete step-by-step answer:

Total number of letters in MISSISSIPPI = 11 letters (4 – S, 4 – I, 2 – P, 1 – M)

Total number of ways of arranging MISSISSIPPI $ = \dfrac{{11!}}{{\left( {4!} \right)\left( {4!} \right)\left( {2!} \right)}}$

Now we need all S to be together

So, if we consider SSSS as 1 block then the remaining number of letters will be MIIIPPI = 8 letters

Number of ways of such arrangement $ = \dfrac{{8!}}{{\left( {4!} \right)\left( {2!} \right)}} = \dfrac{{8!}}{{4! \times 2}}$

Probability that all 4 S’s are together $ = \dfrac{2}{1}$

=$ = \dfrac{{8!}}{{\left( {4!} \right) \times 2}} \times = \dfrac{{\left( {4!} \right)\left( {4!} \right)\left( {2!} \right)}}{{11!}}$

$ = \dfrac{{8!}}{{11 \times 10 \times 9 \times 8!}} \times 4!$

$ = \dfrac{{4 \times 3 \times 2 \times 1}}{{11 \times 10 \times 9}} = \dfrac{4}{{165}}$

Therefore the probability of all ‘S’ are together =$\dfrac{4}{{165}}$

The correct answer is option (A)

Note: To solve such a question we first see what is the probability that four S’s which come consecutively if all the letters of the word MISSISSIPPI are rearranged randomly.

Complete step-by-step answer:

Total number of letters in MISSISSIPPI = 11 letters (4 – S, 4 – I, 2 – P, 1 – M)

Total number of ways of arranging MISSISSIPPI $ = \dfrac{{11!}}{{\left( {4!} \right)\left( {4!} \right)\left( {2!} \right)}}$

Now we need all S to be together

So, if we consider SSSS as 1 block then the remaining number of letters will be MIIIPPI = 8 letters

Number of ways of such arrangement $ = \dfrac{{8!}}{{\left( {4!} \right)\left( {2!} \right)}} = \dfrac{{8!}}{{4! \times 2}}$

Probability that all 4 S’s are together $ = \dfrac{2}{1}$

=$ = \dfrac{{8!}}{{\left( {4!} \right) \times 2}} \times = \dfrac{{\left( {4!} \right)\left( {4!} \right)\left( {2!} \right)}}{{11!}}$

$ = \dfrac{{8!}}{{11 \times 10 \times 9 \times 8!}} \times 4!$

$ = \dfrac{{4 \times 3 \times 2 \times 1}}{{11 \times 10 \times 9}} = \dfrac{4}{{165}}$

Therefore the probability of all ‘S’ are together =$\dfrac{4}{{165}}$

The correct answer is option (A)

Note: To solve such a question we first see what is the probability that four S’s which come consecutively if all the letters of the word MISSISSIPPI are rearranged randomly.