combinatorics
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There are 9 committees with each committee having 20 members and any two committee have 5 members in common then find the minimum number of distinct members in this group.....
Re: combinatorics
Let $5$ members be common for all $9$ committees [Let they be called VIP]
So, the number of non-VIP members of each committee is=$20-5=15$ and, in total=$15 \times 9=135$
Adding the VIP, the number of total members=$135+5=140$
So, the number of non-VIP members of each committee is=$20-5=15$ and, in total=$15 \times 9=135$
Adding the VIP, the number of total members=$135+5=140$
- M. M. Fahad Joy
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Re: combinatorics
Sorry, don't understand. Would you please make understand me?
You cannot cross the sea merely by standing and staring at the water.
- M. M. Fahad Joy
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Re: combinatorics
Thanks a lot.
You cannot cross the sea merely by standing and staring at the water.
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Re: combinatorics
You can get many information about combinatorics hereM. M. Fahad Joy wrote: ↑Sat Feb 17, 2018 9:25 pm
Sorry, don't understand. Would you please make understand me?
- M. M. Fahad Joy
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- Joined:Sun Jan 28, 2018 11:43 pm
Re: combinatorics
Wow... It is like a whole course in Bangla. The administrators should active these forums all the time.
You cannot cross the sea merely by standing and staring at the water.
- M. M. Fahad Joy
- Posts:120
- Joined:Sun Jan 28, 2018 11:43 pm
Re: combinatorics
Here 5 members are common in two committee, not nine.
But why?
But why?
You cannot cross the sea merely by standing and staring at the water.
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Re: combinatorics
You get wrong.It is two committee it is any two committee.