Permutation and combination

Answer :

∵       ⁿC_r = ⁿC_r–1

                   ∴       [n! / {r! (n – r)}] = [n! / {(r – 1)! (n – r + 1)!}]

                   ∴       [1 / {r(r – 1)! (n – r)!}] = [1 / {(r – 1)! (n – r + 1) (n – r)!}]

                   ∴       (1/r) = [1 / (n – r + 1)]    hence 2r – n = 1                       (1)

                   also ⁿP_r = ⁿP_r+1 hence [n! / {(n – r)!}] = [n! / {(n – r – 1)!}]

                   ∴       [1 / (n – r)] = 1               hence n – r = 1                         (2)

                   from (1) & (2) 2r – n + n – r = 2       i.e.     r = 2

                                      and n = 1 + r = 3

                   ∴       n = 3

Answer :

Here 3, 3, 5, 5, are odd digits and 2, 2, 8, 8, 8, are even digits

                   O → odd and E → even other

                   ∴ Total nine digit numbers are [(⁴C₄ × 4!) / (2! 2!)] × [(⁵C₅ × 5!) / (3! 2!)]

                                      = [(1 × 24) / 4] × [(1 × 120) / 12]

                                      = 6 × 20

                                      = 60 ways

Answer :

Here S occurs 3 times, C → 2 times and U, E

                   ∴       Total words = [7! / (3! 2!)]

                   Words with 3 S together = (5! / 2!)

                   ∴       Words with all S not together = [7! / (3! 2!)] – (5! / 2!)

                                                                       = 420 – 60 = 360

Answer :

given: coefficient of 5^th, 6^th, 7^th, term of (1 + x)ⁿ are in A.P.

                   ⇒       ⁿC₄, ⁿC₅, ⁿC₆, respectively are in A.P.

                   ∴       [(ⁿC₄ + ⁿC₆) / 2] = ⁿC₅

                   ∴       [n! / {4! (n – 4)!}] + [n! / {6! (n – 6)!}] = [n! / {5! (n – 5)!}] × 2

                   ∴       [{5! (n – 5)!} / {4! (n – 4)!}] + [{5! (n – 5)!} / {6! (n – 6)!}] = 2

                   ∴       [5 / (n – 4)] + [(n – 5) / 6] = 2

                   ∴       12(n – 4) = 30 + n² – 4n – 5n + 20

                   ∴       n² – 21n + 98 = 0

                   ∴       (n – 7) (n – 14) = 0

                   ∴       n – 7 = 0 or n – 14 = 0

                   ∴       n = 7 or n = 14

                   hence n = 7

Answer :

Given digits: 0, 1, 2, 3, 4, 5, no repetition allowed

                   The required numbers will be of 4, 5 and 6 digits

                   Number of such    4 digit numbers = 3 × ⁵P­₃ = 3 × 5 × 4 × 3 =180

                                                5 digit numbers = ⁶P₅ – ⁵P₄ = 720 – 120 = 600

                                                6 digit numbers = ⁶P₆ – ⁵P₅ = 6! – 5! = 600

                                  ∴ Total numbers      = 180 + 600 + 600

                                                          = 1380