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2023 LMT Fall
1B
2023 Fall Theme p1B
2023 Fall Theme p1B
Source:
December 23, 2023
2023
FAlL
theme
Combo
Problem Statement
Evaluate
(
6
0
)
+
(
6
1
)
+
(
6
4
)
+
(
6
3
)
+
(
6
4
)
+
(
6
5
)
+
(
6
6
)
\dbinom{6}{0}+\dbinom{6}{1}+\dbinom{6}{4}+\dbinom{6}{3}+\dbinom{6}{4}+\dbinom{6}{5}+\dbinom{6}{6}
(
0
6
)
+
(
1
6
)
+
(
4
6
)
+
(
3
6
)
+
(
4
6
)
+
(
5
6
)
+
(
6
6
)
Proposed by Jonathan Liu Solution.
64
\boxed{64}
64
We have that
(
6
4
)
=
(
6
2
)
\dbinom{6}{4}=\dbinom{6}{2}
(
4
6
)
=
(
2
6
)
, so
∑
n
=
0
6
(
6
n
)
=
2
6
=
64
.
\displaystyle\sum_{n=0}^{6} \dbinom{6}{n}=2^6=\boxed{64}.
n
=
0
∑
6
(
n
6
)
=
2
6
=
64
.
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