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Application of multinomial in permutation with repetition:-

⦿ Number of selection/combination of r things out of n things of which p are alike of one kind & q are alike of second kind and remaining (n-p-q) things are all different is given by 
⟹ coefficient of xʳ in (1+x+x²+……..+xᵖ)(1+x+x²+……..+xᑫ){(1+x)(1+x)……… till (n-p-q) times}

= coefficient of +...... + )+...... + )(1

⦿ The number of permutation/arrangement of r things out of n things of which p are alike of one kind & q are alike of second kind and so on……….

= r!.coefficient of

Example:- Let there are
                                p, q ━ 5
                                r, s ━ 4
                                    t ━ 3
                          u, v, w ━ 1
then find the number of permutation/arrangement of 5 letter words selected from given letters
Solution:- 

Using Multinomial theorem:-

Required number of permutations = 5!.coefficient of in = 5!.coefficient of in = 120.coefficient of in = 120.coefficient of in . = 120.coefficient of in = 120= 120= 21462 Answer= 120.coefficient of in

Method(2):       Listing all cases: 

p p p p pp' box5pq q q q q 5qq' boxr r r r 4rr' boxs s s s 4ss' boxt t t3tt' boxu1uu' boxvwv' boxw' box1v 1w S.No.Selection typeNumber of waysof selectionNumber of ways of permutationNumber of words formedof given type12345675-alike²C₁²C₁ × 14-alike +1-different⁴C₁×⁷C₁⁴C₁×⁷C₁×3-alike + 2-alike⁵C₁×⁴C₁⁵C₁×⁴C₁×3-alike + 1-diff.+ 1-diff⁵C₁×⁷C₂⁵C₁×⁷C₂×2-alike + 2-alike1-different⁵C₂×⁶C₁⁵C₂×⁶C₁×2-alike + 1-diff.1-diff. + 1-diff.⁵C₁×⁷C₃⁵C₁×⁷C₃×all different⁸C₅5!⁸C₅×5!required answer will be sum of all terms of this column = 2 + 140 + 200 + 2100 + 1800+ 12500 + 6720 = 21462 Answer

Example:- In how many ways 4 letters of the word ‘PARALLEL’ can be selected & how many 4 letter words can be formed?
Solution:-

L ━ 3        A ━ 2        P ━ 1        R ━ 1        E ━ 1
∴ Required number of ways sof selection:-

= coefficient of in = coefficient of in = cooefficient of in = 2 + 9 + 9 + 2= 22 Answer⦿ Required number of ways of permutation: -= 4!.coefficient of in = 4!. coefficient of in = 4!. coefficient of in = 4!.coefficient of in = = = 286 Answer
Method (2): - Listing all possible cases:L L LL' boxA AA' boxPREP' boxR' boxE' boxS. No.Selection TypeNumber of waysof selectionNumber of waysof permutationNumber of wordsformed of given type12343 - alike + 1 - diff.2 - alike + 2 - alike2 - alike + 1 - diff.+ 1- diff.all different1×⁴C₁1²C₁×⁴C₂⁵C₄1×⁴C₁ײC₁×⁴C₂×⁵C₄×∴ Total number of ways of selection =1×⁴C₁ +1 +²C₁×⁴C₂ +⁵C₄= 4 + 1 + 2×6 + 5= 22 Answer& Total number of 4 letter words = 16 + 6 + 144 + 120 = 288 Answer

Example:- Find the number of 5-letter words that can be formed from the letters given below:-

p, q, r ━ 6
     s, t ━ 5
     u, v ━ 4
     w, y ━ 3
Solution:- 

Method(1):

Required number of 5-letter words:-

= 5!.coefficient of in = 5!.coefficient of in equation (i)equation (ii)equation (iii)equation (iv)Now equation (i): - = =========

∴ Required number of 5-letter words:-
5!.coefficient of x⁵ in
equation(i).equation(ii).equation(iii).equation(iv)

= 5!.coefficient of in = 5!.coefficient of in = 5.coefficient of in = = = 58965 Answer
Method (2): - Listing all possible cases:ppppppp' boxqqqqqqq' boxrrrrrrssssstttttr' boxs' boxt' boxS. No.Selection TypeNumber of waysof selectionNumber of waysof permutationNumber of wordsformed of given type12345 - alike4 - alike + 1 - diff3 - alike + 2 - alike⁷C₁×⁸C₁⁵C₁uuuuvvvvwwwyyyu' boxv' boxw' boxy' box6 6 6 5 5 4 4 3 3567all different3 - alike + 1 - diff. + 1 - diff.2 - alike + 2 - alike + 1 - diff.2 - alike + 1 - diff. + 1 - diff. + 1 - diff.⁹C₁×⁸C₁⁹C₁×⁸C₂⁹C₂×⁷C₁⁹C₁×⁸C₃⁹C₅⁵C₁×1 = 5⁷C₁×⁸C₁×5 = 280⁹C₁×⁸C₁×10 = 720⁹C₁×⁸C₂×20 = 5040⁹C₂×⁷C₁×30 = 7560⁹C₁×⁸C₃×60 = 30240⁹C₅×120 = 15120∴ Required number of 5-letter words = 5 + 280 + 720 + 5040 + 7560 + 30240 + 15120 = 58965 Answer

Method(3):

            p, q, r ━ 6
                  s, t ━ 5
                  u, v ━ 4
                  w, y ━ 3

Counting favourable & unfavourable cases:-

99999= when all lettersavailable at least 5-timesunfavourable cases: -u u u u uv v v v vw w w w wy y y y yunfavourable cases: -as: - w w w wq= 58965 Answer

(Q) consider a word ‘EXAMINATION’. In how many ways 4 letters of this word can be selected & how many 4-letter words can be formed from this word?
Solution:-

using multinomial theorem:-

                    A ━ 2
                     I ━ 2
                    N ━ 2
E, X, M, T, O ━ 1

∴ required number of selection  

= coefficient of x⁴ in (1 + x + x²)³ (1 + x)⁵
= coefficient of x⁴ in (1 +  x² + x⁴ + 2x + 2x³ + 2x²) (1 + x + x²) (1 + x)⁵
=  coefficient of x⁴ in (1 + 2x + 3x² + 2x³ + x⁴) (1 + x + x²) (1 + x)⁵
= coefficient of x⁴ in
(1 + 2x + 3x² + 2x³ +  x⁴
    +  x  +  2x² + 3x³ + 2x⁴  +  x⁵
             +   x²  + 2x³ + 3x⁴  +  2x⁵ + x⁶).(1 + x)⁵
=  coefficient of x⁴ in (1 + 3x + 6x² + 7x³ +  6x⁴)*(1 + x)⁵
= ⁵C₄ + 3×⁵C₃ + 6×⁵C₂ + 7×⁵C₁ + 6×1
= 5 + 3×10 + 6×10 + 7×5 + 6
136 Answer

⦿ Now Required number of 4 – letter words:- 

= 4!.coefficient of in = 4!.coefficient of in = 4!.coefficient of in = 4!.coefficient of in = 4!.coefficient of in = = = = = 2454 Answer
Method (2): - Listing all possible cases:S. No.Selection TypeNumber of waysof selectionNumber of waysof permutationNumber of wordsformed of given type12344 - alike3 - alike + 1 - diff2 - alike + 2 - alike0052 - alike + 1 - diff. + 1 - diff.All different187561680A AA' box I' box N' box E' box X' box M' box T' box O' boxI IN NEXMTO2 2 2 1 1 1 1 1 0000

Hence from above table ⟹
Required number of selection = 3 + 63 + 70
                                                       = 136 Answer

& Required number of 4-letter words = 18 + 756 + 1680
                                                                     = 2454 Answer