The number of ways in which an arrangement of | vedclass

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Question

(C) The word $PROPORTION$ contains $10$ letters: $P(2), R(2), O(3), I(1), T(1), N(1)$. There are $6$ distinct types of letters: ${P, R, O, I, T, N}$.

Vedclass · Accepted Answer

$758$

Vedclass · Answer

(C) The word $PROPORTION$ contains $10$ letters: $P(2), R(2), O(3), I(1), T(1), N(1)$. There are $6$ distinct types of letters: ${P, R, O, I, T, N}$. We need to arrange $4$ letters.Case $(i)$: All $4$ letters are different.Number of ways to select $4$ distinct letters from $6$ types is $^6C_4 = 15$.Arrangements $= 15 	imes 4! = 15 	imes 24 = 360$.Case $(ii)$: $2$ letters are alike and $2$ are different.Select $1$ pair from ${P, R, O}$ in $^3C_1 = 3$ ways. Select $2$ different letters from the remaining $5$ types in $^5C_2 = 10$ ways.Number of selections $= 3 	imes 10 = 30$.Arrangements $= 30 	imes \frac{4!}{2!} = 30 	imes 12 = 360$.Case $(iii)$: $2$ letters are alike of one kind and $2$ are alike of another kind.Select $2$ pairs from ${P, R, O}$ in $^3C_2 = 3$ ways.Arrangements $= 3 	imes \frac{4!}{2!2!} = 3 	imes 6 = 18$.Case $(iv)$: $3$ letters are alike and $1$ is different.Select the set of $3$ alike letters $(O)$ in $1$ way. Select $1$ different letter from the remaining $5$ types in $^5C_1 = 5$ ways.Arrangements $= 5 	imes \frac{4!}{3!} = 5 	imes 4 = 20$.Total arrangements $= 360 + 360 + 18 + 20 = 758$.

The number of ways in which an arrangement of $4$ letters of the word $PROPORTION$ can be made is

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