There are m apples and n oranges to be placed in a line such that the two extreme fruits being both oranges . Let P denotes the numbe of arrangements if the fruits of the same species are different and Q the corresponding figure when the fruits of the same species are alike, then the ratio P / Q has the value equal to :

Question

There are m apples and n oranges to be placed in a line such that the two extreme fruits being both oranges . Let P denotes the numbe of arrangements if the fruits of the same species are different and Q the corresponding figure when the fruits of the same species are alike, then the ratio P / Q has the value equal to : (A)  n P 2 ⋅ m P m ⋅( n -2) ! (B)  m P2 ⋅ n Pn ⋅(n-2) ! (C)  n P 2 ⋅ n P n ⋅( m -2) ! (D) none

Tardigrade Admin · Accepted Answer

P = n C 2 ⋅ 2 !( m + n -2) ! ; Q =(( m + n -2) !/( n -2) ! m !)

$^{n} P_{2} \cdot^{m} P_{m} \cdot (n - 2)!$

$^{m} P_{2} \cdot^{n} P_{n} \cdot (n - 2)$ !

$^{n} P_{2} \cdot^{n} P_{n} \cdot (m - 2)$ !

none

$P =^{n} C_{2} \cdot 2! (m + n - 2)!; Q = \frac{( m + n - 2 )!}{( n - 2 )! m !}$

nP2​⋅mPm​⋅(n−2)!

mP2​⋅nPn​⋅(n−2) !

nP2​⋅nPn​⋅(m−2) !

none

P=nC2​⋅2!(m+n−2)!;Q=(n−2)!m!(m+n−2)!​

$^{n} P_{2} \cdot^{m} P_{m} \cdot (n - 2)!$

$^{m} P_{2} \cdot^{n} P_{n} \cdot (n - 2)$ !

$^{n} P_{2} \cdot^{n} P_{n} \cdot (m - 2)$ !

$P =^{n} C_{2} \cdot 2! (m + n - 2)!; Q = \frac{( m + n - 2 )!}{( n - 2 )! m !}$