There are six mathematics books, three geography books and seven history books on a bookshelf. Number of different ways in which one can select four books, so that each selection must have atleast one mathematics book, (Assume that books of the same subject are different.)

Question

There are six mathematics books, three geography books and seven history books on a bookshelf. Number of different ways in which one can select four books, so that each selection must have atleast one mathematics book, (Assume that books of the same subject are different.) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

6 maths books, 3 geography books and 7 history books. To select four books such that atlest one mathematics book is there. = Total ways - ways when maths books are not there. = 16 C 4- 10 C 4=1820-210=1610

Q. There are six mathematics books, three geography books and seven history books on a bookshelf. Number of different ways in which one can select four books, so that each selection must have atleast one mathematics book, (Assume that books of the same subject are different.)

$6$ maths books, $3$ geography books and $7$ history books.
To select four books such that atlest one mathematics book is there.
$=$ Total ways - ways when maths books are not there.
$=^{16} C_{4} -^{10} C_{4} = 1820 - 210 = 1610$

Q. There are six mathematics books, three geography books and seven history books on a bookshelf. Number of different ways in which one can select four books, so that each selection must have atleast one mathematics book, (Assume that books of the same subject are different.)

6 maths books, 3 geography books and 7 history books. To select four books such that atlest one mathematics book is there. = Total ways - ways when maths books are not there. =16C4​−10C4​=1820−210=1610

$6$ maths books, $3$ geography books and $7$ history books.
To select four books such that atlest one mathematics book is there.
$=$ Total ways - ways when maths books are not there.
$=^{16} C_{4} -^{10} C_{4} = 1820 - 210 = 1610$