1 votes 1 votes The number of seven digit integers possible with sum of the digits equal to 11 and formed by using the digits 1, 2 and 3 only are Mathematical Logic combinatory + – rajoramanoj asked Jan 17, 2018 rajoramanoj 273 views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply joshi_nitish commented Jan 17, 2018 reply Follow Share the answer will be, $[x^{11}]$ in $(x+x^2+x^3)^7$ $[x^{11}]$ in $x^7(1-x^3)^7(1-x)^{-7}$ $[x^4]$ in $(1-x^3)^7(1-x)^{-7}$ $=\binom{-7}{4}+\binom{7}{1}\binom{-7}{1}$ $=210 - 49 = 161$ 3 votes 3 votes Ashwin Kulkarni commented Jan 17, 2018 reply Follow Share Case 1 : 4 one's , 2 two's, 1 three $= \frac{7!}{4!*2!} = 105$ Case 2 : 5 one's, 2 three's $= \frac{7!}{5! * 2!} = 21$ Case 3 : 3 one's, 4 two's $= \frac{7!}{4!*3!} = 35$ Total $105 + 21 + 35 = 161$ 2 votes 2 votes Please log in or register to add a comment.