The number of ways in which an arrangement of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

The number of ways of dividing $52$ cards amongst four players so that three players have $17$ cards each and the fourth player has just one card,is

Each section in the first year of a plus $2$ course has exactly $40$ students. If there are $5$ sections,in how many ways can a set of $4$ student representatives be selected,$1$ from each section?

$A$ person goes in for an examination in which there are four papers with a maximum of $m$ marks from each paper. The number of ways in which one can get $2m$ marks is

If $n$ and $r$ are two positive integers such that $n \ge r,$ then $^nC_{r-1} + ^nC_r = $

There are $6$ items in column $A$ and $6$ items in column $B$. $A$ student is asked to match each item in column $A$ with an item in column $B$. How many possible (correct or incorrect) answers are there to this question?

Vedclass Test Series

Exam Paper Generator

Online Exam Module