Question

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(9) (5 pts) How many solutions are there to the inequality 11 +*2+I3 <11, where 21, 22, and 13 are non-negative integers? Hint: Introduce an auxiliary variable In such that 21+12+13+14 = 11. (10) (5 pts) How many terms are there in the expansion of (2x + 3y + 5z)50?

Answer 1

9 11 761 are Solution x + x2 + 23 and 262, 263 non negative integers, 4,7 l 2271 213 > 24, + 262 + x3 +24 11 (-from question) - 7 9 2 1 + x + 1 +242 + 1+ x3 + 1+ x + 1 + xús = 11 ri + x₂ + x3 +216 { x s xixo solution - 17+4-1) x = 3 = 0,22 ? = 7 x47 No. of Cetro (4-1) 10 c3 120 Answer 10 X 9*8 3 X 2X1 solutions are Number of 12o solution 10 (22 + 3y + 5z) 5° р ४ (2x) (3y)? ( 5z) Tn n! = Pt q tr=n=50 P! q 181 solution n+3-1 с For p+ q to 7 = 5o is 3-) (OY) nt2 52ca C₂ 52 X 51 2 x 1 26 X 51 - 1 326 Terms Answer

metho For 50 solution 10 (2% + 3y + 5z) put (2x+3y) Take In=50 (a+ x = (n+1) terons ť Know, + n ( + 5z) (n+1) terms n-a c. (X)^ ( 4z)- locaan) => all the this are terms Form of n-a cil Dx+34) 2) 9. (52) a = 0 + 1 + 2 + inti) + 2 + 3+ - tn (nu) using Identity 2 - terms (n+1)(n+2) no al 2 put n=50 terms 51 X 52 - no. of 1326 terms. 2