# Answer to Question #4398 in Combinatorics | Number Theory for Bliss

Question #4398

Find the number of ways of forming a group of 2k people from n couple, where k,n are elements of N with 2k≤n in each case: i) k couples are in such a group, ii) no couples are included in such a group, iii) at least one couple is included in such a group, and iv) exactly two couples are included in such a group.

Expert's answer

i) k couples are in such a groupWe should pick up k couples from n

ii) no couples are included insuch a group

at first time we can pick nvariants

second- n-2(cannot pick 1 couple one member of whitchis picked)

…

So ans n(n-2)...(n-2k)/(2k)!

iii) at least one couple isincluded in such a group

C

n(n-2)..(n-2k)/(2k)! it does not fitus.

So ans:C

iv) exactly two couples are included in such a group.

C

C

so ans. C

ii) no couples are included insuch a group

at first time we can pick nvariants

second- n-2(cannot pick 1 couple one member of whitchis picked)

…

So ans n(n-2)...(n-2k)/(2k)!

iii) at least one couple isincluded in such a group

C

^{2k}_{2n}-all variants to pick up 2k from 2nn(n-2)..(n-2k)/(2k)! it does not fitus.

So ans:C

^{2k}_{2n }- n(n-2)..(n-2k)/(2k)!iv) exactly two couples are included in such a group.

C

^{2}_{n}-so many ways areto pick 2 couplesC

^{2k-4}_{2n-4}-we pick from 2n-4people 2k-4 places in our groupso ans. C

^{2}_{n}*C^{2k-4}_{2n-4}Need a fast expert's response?

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