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6.Permutation and Combination
medium
All possible two factors products are formed from numbers $1, 2, 3, 4, ...., 200$. The number of factors out of the total obtained which are multiples of $5$ is
A
$5040$
B
$7180$
C
$8150$
D
None of these
Solution
(b) The total number of two factor products ${ = ^{200}}{C_2}$.
The number of numbers from $1$ to $200$ which are not multiples of $5$ is $160$.
Therefore total number of two factor products which are not multiple of $5$ is $^{160}{C_2}$.
Hence the required number of factors ${ = ^{200}}{C_2}{ – ^{160}}{C_2} = 7180$.
Standard 11
Mathematics