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Math Invitational for Girls
2018 MIG
21
21
Part of
2018 MIG
Problems
(1)
2018 Individual #21
Source:
1/13/2023
Find the sum:
11
×
(
2
0
)
+
10
×
(
3
1
)
+
9
×
(
4
2
)
+
⋯
+
2
×
(
11
9
)
+
(
12
10
)
11 \times \dbinom20 + 10 \times \dbinom31 + 9 \times \dbinom42 + \cdots + 2 \times \dbinom{11}9 + \dbinom{12}{10}
11
×
(
0
2
)
+
10
×
(
1
3
)
+
9
×
(
2
4
)
+
⋯
+
2
×
(
9
11
)
+
(
10
12
)
Where
(
n
r
)
\tbinom{n}{r}
(
r
n
)
is combination function given by
n
!
r
!
(
n
−
r
)
!
\tfrac{n!}{r!(n-r)!}
r
!
(
n
−
r
)!
n
!
<
s
p
a
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c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
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′
>
(
A
)
<
/
s
p
a
n
>
351
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
841
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
901
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
991
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
1001
<span class='latex-bold'>(A) </span> 351\qquad<span class='latex-bold'>(B) </span> 841\qquad<span class='latex-bold'>(C) </span> 901\qquad<span class='latex-bold'>(D) </span> 991\qquad<span class='latex-bold'>(E) </span> 1001
<
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a
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>
(
A
)
<
/
s
p
an
>
351
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
841
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
901
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
991
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
1001
2018 Individual