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Total number of socks = 5+4+3 = 12 4 to be selected So number of ways of selecting = 12$C$4 = 495 ways

Is this correct?

Ayush Sambher
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    This is correct if the socks are distinguishable. However, if you consider selecting any 4 blue socks to be one "way" (regardless of precisely which 4 blue socks you chose), then the answer will be different. – angryavian Sep 07 '21 at 17:57
  • No, that can't be correct. If it were correct, the whole point of designating that some of the socks are blue, some are green, some are red would become meaningless. The only way that the question makes sense is to assume that all socks of the same color are indistinguishable from each other. – user2661923 Sep 07 '21 at 17:57
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    That is correct assuming that he can tell the difference between each of the blue socks... that there is a "first" blue sock and a "second" blue sock etc... If all he cares about is the quantity of each color sock and doesn't care about specifically which of each type of sock it was... then this is incorrect. – JMoravitz Sep 07 '21 at 17:57
  • Since the numbers involved are so small, I suggest forgoing any attempt at elegance, and considering the following $5$ cases: $k$ blue socks are selected, where $k \in {0,1,2,3,4}.$ For each case, you may have subcases to manually enumerate. – user2661923 Sep 07 '21 at 18:00

3 Answers3

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There are so few cases that we can just list artistically in the order blue-red-green in descending numbers

$\displaylines {400\\310\; 301\\220\; 211\; 202\\130\;121\;112\;103\\040\;031\;022\;013}$

true blue anil
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You could have used $^{12}C_4$ if all items were different, now now.

Now, you can have the following cases.

Case I: All $4$ socks of the same color. $2$ ways. (either blue or red)

Case II: $3$ socks of the same color, $4$th of the different color. $^3C_1\times^2C_1=6$ ways.

Case III: $2$ socks of one color. Other $2$ of other color. i.e. we are picking $2$ colors from which $2$-$2$ socks each would be picked. $^3C_2=3$ ways.

Case IV: $2$ socks of one color. Other $2$ of two other colors. $^3C_1=3$ ways.

Total$=2+6+3+3=14$ ways.

aarbee
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This is a Stars and Bars problem, which is discussed here and here.

Let $x_1, x_2, x_3$ denote the number of blue, red, and green socks selected, respectively.

The number of non-negative integer solutions to
$x_1 + x_2 + x_3 = 4$

is $~\displaystyle \binom{4 + 2}{2} = \binom{6}{2} = 15.$

The only possible violation to the constraint that
$x_1 \leq 5, ~x_2 \leq 4, ~x_3 \leq 3$
is $(x_1,x_2,x_3) = (0,0,4).$

Therefore, the overall enumeration is $15 - 1 = 14.$

user2661923
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