The correct answer is d) 1020
In the given numbers 1, 3,5,7,…, 147, 148, 151the numbers which are multiple of 5 are 5, 15, 25, 35,…, 145 which are an arithmetic sequence .
\(T_n=a+(n-1)d\)
\(=145=5+(n-1)10\)
\(\therefore n=15\)
and if total number of terms in the given sequence is m ,then
\(151=1+(m-1)2\)
\(\therefore m=76\)
so, the number of ways in which product is a multiple of 5=(both two numbers from 5, 15 , 25 ,.....,145) or (one number from 5, 15, 25 ,.....145 and one from remaining)
\(=^{15}C_2+^{15}C_1\times ^{76-15}C_1\)
\(=^{15}C_2+^{15}C_1\times ^{61}C_1\)
\(={ 15\times14\over 2}+15\times{61}\)
\(=105+915
=1020\)
