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Two numbers are chosen from 1, 3, 5, 7,… 147, 149 and 151 and multiplied together in all possible ways. The number of ways which will give us the product a multiple of 5, is

Options:

 (a) 75                                                              

(b) 1030

(c) 95                                                               

(d) 1020
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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) 

\(=^{15}C_2+^{15}C_1\times ^{76-15}C_1\) 

\(=^{15}C_2+^{15}C_1\times ^{61}C_1\) 

\(={ 15\times14\over 2}+15\times{61}\) 

\(=105+915 =1020\)

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