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showing 1-2 of 2 (answers : 2, comments : 0),   sorted newest to the oldest
Question
A boy has $3$ library tickets and $8$ books of his interest in the library of these $8,$ he does not want to borrow maths part $2$ unless maths part $1$ is borrowed? In how many ways can he choose the three books to be borrowed?
(a) $41$
(b) $51$
(c) $52$
(d) $61$
 
(1) (0) Comment Answer
permutations & combinations
2016-12-21 12:17:28 
Keyur thakkar
2 Answers
Case 1: 3 books without including maths part 1 and part 2
$6\text{C}3=20$ ways

Case 2: (maths part 1) and (two books)
$7\text{C}2=21$ ways

Required number of ways
$=20+21=41$
 
(0) (0) Comment
2016-12-21 12:48:22 
Jay Patel (Senior Maths Expert, careerbless.com)
Number of ways in which three books can be selected without any conditions
$=8\text{C}3=56$.

Now find out number of ways in which three books can be selected in which maths part $2$ must appear and part $1$ must not appear. For this, remove maths part $1$ and part $2$ first and select two books from the remaining $6$ books. This can be done in $6\text{C}2=15$ ways.

Required number of ways
$=56-15=41$
 
(0) (0) Comment
2016-12-21 12:43:40 
jiju (Junior Maths Expert, careerbless.com)
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