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Compute:
\[\binom50+\binom51+\binom62+\binom71+\binom83+\binom92+\binom{10}4+\binom{11}3\]
using Pascal's Identity.

 Apr 29, 2015

Best Answer 

 #1
avatar+26396 
+10

\[(50)+(51)+(62)+(71)+(83)+(92)+(104)+(113)\]=495

see: http://web2.0calc.com/#binom(5,0)+binom(5,1)+binom(6,2)+binom(7,1)+binom(8,3)+binom(9,2)+binom(10,4)+binom(11,3) and push the "=" Button

 

 (00)=1(10)=1(11)=1(20)=1(21)=2(22)=1(30)=1(31)=3(32)=3(33)=1(40)=1(41)=4(42)=6(43)=4(44)=1(50)=1(51)=5(52)=10(53)=10(54)=5(55)=1(60)=1(61)=6(62)=15(63)=20(64)=15(65)=6(66)=1(70)=1(71)=7(72)=21(73)=35(74)=35(75)=21(76)=7(77)=1(80)=1(81)=8(82)=28(83)=56(84)=70(85)=56(86)=28(87)=8(88)=1(90)=1(91)=9(92)=36(93)=84(94)=126(95)=126(96)=84(97)=36(98)=9(99)=1(100)=1(101)=10(102)=45(103)=120(104)=210(105)=252(106)=210(107)=120(108)=45(109)=10(1010)=1(110)=1(111)=11(112)=55(113)=165(114)=330(115)=462(116)=462(117)=330(118)=165(119)=55(1110)=11(1111)=1

 

 (50)+(51)+(62)+(71)+(83)+(92)+(104)+(113)=1+5+15+7+56+36+210+165=495

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 Apr 30, 2015
 #1
avatar+26396 
+10
Best Answer

\[(50)+(51)+(62)+(71)+(83)+(92)+(104)+(113)\]=495

see: http://web2.0calc.com/#binom(5,0)+binom(5,1)+binom(6,2)+binom(7,1)+binom(8,3)+binom(9,2)+binom(10,4)+binom(11,3) and push the "=" Button

 

 (00)=1(10)=1(11)=1(20)=1(21)=2(22)=1(30)=1(31)=3(32)=3(33)=1(40)=1(41)=4(42)=6(43)=4(44)=1(50)=1(51)=5(52)=10(53)=10(54)=5(55)=1(60)=1(61)=6(62)=15(63)=20(64)=15(65)=6(66)=1(70)=1(71)=7(72)=21(73)=35(74)=35(75)=21(76)=7(77)=1(80)=1(81)=8(82)=28(83)=56(84)=70(85)=56(86)=28(87)=8(88)=1(90)=1(91)=9(92)=36(93)=84(94)=126(95)=126(96)=84(97)=36(98)=9(99)=1(100)=1(101)=10(102)=45(103)=120(104)=210(105)=252(106)=210(107)=120(108)=45(109)=10(1010)=1(110)=1(111)=11(112)=55(113)=165(114)=330(115)=462(116)=462(117)=330(118)=165(119)=55(1110)=11(1111)=1

 

 (50)+(51)+(62)+(71)+(83)+(92)+(104)+(113)=1+5+15+7+56+36+210+165=495

heureka Apr 30, 2015

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