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If-I-have-10-people-and-I-want-to-create-pairings-so-that-every-person-is-matched-with-every-other-person-at-least-once-how-many-unique-pairings-will-there-be-in-total-




Question Number 224261 by Tawa11 last updated on 30/Aug/25
If I have 10 people and I want to create  pairings, so that every person is matched  with every other person at least once, how  many unique pairings will there be in total?
If I have 10 people and I want to create
pairings, so that every person is matched
with every other person at least once, how
many unique pairings will there be in total?
Commented by mr W last updated on 31/Aug/25
no clear enough what you are asking.  do you ask how many ways to select  a pair from 10 people?  then answer is C_2 ^(10) =45 ways.    or do you ask how many ways to   buid 5 pairs from 10 people?  then answer is ((10!)/((2!)^5 ×5!))=945 ways.
$${no}\:{clear}\:{enough}\:{what}\:{you}\:{are}\:{asking}. \\ $$$${do}\:{you}\:{ask}\:{how}\:{many}\:{ways}\:{to}\:{select} \\ $$$${a}\:{pair}\:{from}\:\mathrm{10}\:{people}? \\ $$$${then}\:{answer}\:{is}\:{C}_{\mathrm{2}} ^{\mathrm{10}} =\mathrm{45}\:{ways}. \\ $$$$ \\ $$$${or}\:{do}\:{you}\:{ask}\:{how}\:{many}\:{ways}\:{to}\: \\ $$$${buid}\:\mathrm{5}\:{pairs}\:{from}\:\mathrm{10}\:{people}? \\ $$$${then}\:{answer}\:{is}\:\frac{\mathrm{10}!}{\left(\mathrm{2}!\right)^{\mathrm{5}} ×\mathrm{5}!}=\mathrm{945}\:{ways}. \\ $$
Commented by TonyCWX last updated on 31/Aug/25
You′re missing one person.  Notice that it′s “10 people and I ”.
$${You}'{re}\:{missing}\:{one}\:{person}. \\ $$$${Notice}\:{that}\:{it}'{s}\:“\mathrm{10}\:{people}\:\boldsymbol{{and}}\:\boldsymbol{{I}}\:''. \\ $$
Commented by mr W last updated on 31/Aug/25
Sorry, my english is not so good.  When you say “I have 3 dogs and I  want to feed them.” According to  my understanding, you just feed  the 3 dogs, not including yourself.   But according to you, 4 persons   should be fed, because one persion   is missing, since it′s “3 dogs and I”.  please recheck!
$${Sorry},\:{my}\:{english}\:{is}\:{not}\:{so}\:{good}. \\ $$$${When}\:{you}\:{say}\:“\underline{{I}\:{have}\:\mathrm{3}\:{dogs}\:{and}\:{I}} \\ $$$$\underline{{want}\:{to}\:{feed}\:{them}.}''\:{According}\:{to} \\ $$$${my}\:{understanding},\:{you}\:{just}\:{feed} \\ $$$${the}\:\mathrm{3}\:{dogs},\:{not}\:{including}\:{yourself}.\: \\ $$$${But}\:{according}\:{to}\:{you},\:\mathrm{4}\:{persons}\: \\ $$$${should}\:{be}\:{fed},\:{because}\:{one}\:{persion}\: \\ $$$${is}\:{missing},\:{since}\:{it}'{s}\:“\mathrm{3}\:{dogs}\:\boldsymbol{{and}}\:\boldsymbol{{I}}''. \\ $$$${please}\:{recheck}! \\ $$
Commented by fantastic last updated on 31/Aug/25
The language of question   should not be misleading
$${The}\:{language}\:{of}\:{question}\: \\ $$$${should}\:{not}\:{be}\:{misleading} \\ $$$$ \\ $$
Commented by Tawa11 last updated on 31/Aug/25
If I have 10 people and want to create pairings where every person is matched with every other person at least once, how many unique pairings will there be?"
Commented by Tawa11 last updated on 31/Aug/25
Sir mrW,  with what you stated, the   ^(10) C_2   =  45  is correct sir. I want to understand the  grammar too, that is why I sent it here.
$$\mathrm{Sir}\:\mathrm{mrW}, \\ $$$$\mathrm{with}\:\mathrm{what}\:\mathrm{you}\:\mathrm{stated},\:\mathrm{the}\:\:\overset{\mathrm{10}} {\:}\mathrm{C}_{\mathrm{2}} \:\:=\:\:\mathrm{45} \\ $$$$\mathrm{is}\:\mathrm{correct}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{want}\:\mathrm{to}\:\mathrm{understand}\:\mathrm{the} \\ $$$$\mathrm{grammar}\:\mathrm{too},\:\mathrm{that}\:\mathrm{is}\:\mathrm{why}\:\mathrm{I}\:\mathrm{sent}\:\mathrm{it}\:\mathrm{here}. \\ $$
Commented by TonyCWX last updated on 01/Sep/25
Sorry sir.  My understanding was bad this time.
$${Sorry}\:{sir}. \\ $$$${My}\:{understanding}\:{was}\:{bad}\:{this}\:{time}. \\ $$

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