The number of arrangements of the letters of | vedclass

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(A) The word $SATAYPAUL$ has $9$ letters: $S, A, T, A, Y, P, A, U, L$. There are $3$ $A$'s and $6$ other letters $(S, T, Y, P, U, L)$. Total letters =

Vedclass · Accepted Answer

$(5!)^2$

Vedclass · Answer

(A) The word $SATAYPAUL$ has $9$ letters: $S, A, T, A, Y, P, A, U, L$. There are $3$ $A$'s and $6$ other letters $(S, T, Y, P, U, L)$. Total letters = $9$. The middle position is the $5$th position. Consonants are $S, T, Y, P, L$ ($5$ consonants). First,we place the middle consonant in the $5$th position. There are $5$ choices for this. Remaining letters are $8$ ($3$ $A$'s and $5$ other letters). Let the remaining $5$ letters be $X_1, X_2, X_3, X_4, X_5$. We arrange these $5$ letters in $5!$ ways. There are $6$ gaps created by these $5$ letters (including ends): $\_ X_1 \_ X_2 \_ X_3 \_ X_4 \_ X_5 \_$. However,the middle position is already occupied by a consonant. Let the arrangement be $L_1 L_2 L_3 L_4 C L_6 L_7 L_8 L_9$. We need to place $3$ $A$'s in the remaining $8$ positions such that no two $A$'s are together. Using the gap method for the $6$ non-$A$ letters,there are $7$ possible gaps. Since the middle position is fixed as a consonant,we exclude the gap adjacent to the middle position if it violates the condition. Total arrangements = $5 	imes 5! 	imes \binom{6}{3} = 5 	imes 120 	imes 20 = 12000 = (5!)^2$.

The number of arrangements of the letters of the word $SATAYPAUL$ such that no two $A$ are together and the middle letter is a consonant,is

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