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In a plant there are 3 units and each unit has 2 identical pumps making it to be 6 pumps in total. How to calculate the number of combinations where both pumps failed on any of the units AND an additional 1 pump on any of the other units (I.e. 3 total failures with 2 on the same unit and 1 on the other unit)?

If we just need to find out the combinations of 3 pumps failing out of the 6 pumps then we can use 6C3, which would give 20. However, now the challenge is to make sure that any one of the unit to have 2 failed pumps and any other unit with 1 failed pump. The possible combinations are illustrated in the attached image and as shown in the image, the answer for this problem should be 12.

Possible combinations

tharma viswa
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    What does "number of combinations" mean here? Moreover, not sure this deals with group theory. –  Aug 22 '22 at 21:53
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    Welcome to MathSE. When you pose a question here, it is expected that you include your own thoughts on the problem. Please edit your question to show what you have attempted and to explain where you are stuck so that you receive responses that address the specific difficulties you are encountering. – N. F. Taussig Aug 22 '22 at 21:57
  • Hint: The constraints of the problem effectively make the $3$ pumps distinguishable from each other. Therefore, the relevant question is : how many ways are there of ordering the $3$ distinguishable pumps? – user2661923 Aug 22 '22 at 22:24

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You need not at all go to the lengths you have gone for solving it.

It can be solved almost mechanically by multiplying combinations

[Choose unit with $2$ failures]$\times$[Choose unit with $1$ failure]$\times$[Choose failed pump]

Thus $^3C_1 \times ^2C_1 \times ^2C_1 = 12$

true blue anil
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  • Thank you Anil, but no matter how many times I tried I could not get 12 combination as the image above. I think what makes this problem different is there are 3 isolated groups and sets of same size but I can fill them in different ways. – tharma viswa Aug 25 '22 at 14:01
  • So sorry for misreading figures, have corrected !$;;$ :) – true blue anil Aug 25 '22 at 14:31