Question Number 103716 by mr W last updated on 16/Jul/20

Commented bymr W last updated on 16/Jul/20

after thinking about it for two days,  i think i have found the general  solution which is simply  C_m ^(n−m+1) .  please give a try, prove or disprove it!

Commented byprakash jain last updated on 16/Jul/20

s_1  s_2   s_3   s_4  ...s_n   if we draw m bars and select  student to the left of bar we have  total n postion for bars.  s_1 ∙ s_2 ∙  s_3   ∙s_4  ∙...∙s_n   So for example 6,2 case  s∣ s s s s∣ s  We need m bars,  the first bar can have 1 or more  students to the left. The remaining  bars can only have 2 or more  student to the left of bar.  0 or more student can be present  after last bar.  Σ_(i=1) ^n x(Σ_(i=2) ^n x^2 )^(m−1) Σ_(i=0) ^n x^i   ((x(1−x^n ))/(1−x))×((x^(2(m−1)) (1−x^(n−1) )^(m−1) )/((1−x)^(m−1) ))×(((1−x^(n+1) ))/(1−x))  m≥1  =x^(2m−1) ×(1/((1−x)^(m+1) ))  =x^(2m−1) Σ_(i=0) ^∞ C_i ^(m+i) x^i   sum of student must be n  2m−1+i=n⇒i=n+1−2m  Required result  C_(n+1−2m) ^(m+n+1−2m) =C_(n+1−2m) ^(n−m+1)   =C_((n−m+1)−(n+1−2m)) ^(n−m+1) =C_m ^(n−m+1)   Please feedback.

Commented bymr W last updated on 16/Jul/20

correct sir!

Answered by mr W last updated on 17/Jul/20

this is my solution    ★ stand for m selected students  ♦⧫ stand for places for other students    ♦★⧫★⧫★⧫★⧫...⧫★⧫★◊    ◊ may be empty or occupied by any  number of students  ⧫ must be occupied by at least one  student    we have two ◊ places, generating  function of each one is  1+x+x^2 +x^3 +...=(1/(1−x))    we have (m−1) ⧫ places, generating  function of each one is  x+x^2 +x^3 +...=(x/(1−x))    the total number of students who  occupy the ◊ and ⧫ places is (n−m).    the number of ways to occupy these  places is the coefficient of the x^(n−m)   term of the following generating  function  (1+x+x^2 +x^3 +...)^2 (x+x^2 +x^3 +...)^(m−1)   =((1/(1−x)))^2 ((x/(1−x)))^(m−1) =(x^(m−1) /((1−x)^(m+1) ))  =x^(m−1) Σ_(k=0) ^∞ C_m ^(k+m) x^k   m−1+k=n−m ⇒k+m=n−m+1  ⇒the coefficient of x^(n−m)  term is  therefore C_m ^(n−m+1)  which is the  answer we need.