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Q.
Number of ways of selecting $2$ numbers form integers $1$ to $100$ if their sum is divisible by $5$ are
Permutations and Combinations
Solution:
The numbers can be categorized as
(a) $5,10,15, \ldots, 100 \rightarrow 5 \lambda$
(b) $1,6,11, \ldots, 96 \rightarrow 5 \lambda+1$
(c) $2,7,12, \ldots, 97 \rightarrow 5 \lambda+2$
(d) $3,8,13, \ldots, 98 \rightarrow 5 \lambda+3$
(e) $4,9,14, \ldots, 99 \rightarrow 5 \lambda+4$
Now, sum of two numbers is divisible by 5 then we can have following cases
(i) both the numbers from the list (a) - number of selection ways are ${ }^{20} C_{2}$
(ii) one number from list (b) and other from list (e) - number of selection ways are ${ }^{20} C_{1} \times{ }^{20} C_{1}$
(iii) one number from list (c) and other from list (d) - number of selection ways are ${ }^{20} C_{1} \times{ }^{20} C_{1}$
Hence, total number of ways are
${ }^{20} C_{2}+{ }^{20} C_{1} \times{ }^{20} C_{1}+{ }^{20} C_{1} \times{ }^{20} C_{1}$
$=990$