Permutations and Combinations

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a month ago

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R
Raj Pandey

Contributor-Level 9

0 Red, 1 Red, 2 Red, 3 Red
Number of ways = ?C? + ?C?.?C? + ?C?.?C? + ?C?.?C? = 35+175+210+70=490

New answer posted

a month ago

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V
Vishal Baghel

Contributor-Level 10

A → 5Q; B → 5Q; C → 5QA
A?, A?, A?, A?, A?; B?, B?, B?, B?, B?; C?, C?, C?, C?, C?
A?A?A?B?C? → ?C? * ?C? * ?C? = 750
A?A?B?B?C? → ³C? * ?C? * ?C? = 1500
∴ Total = 2250

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a month ago

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A
alok kumar singh

Contributor-Level 10

6 35 C r = k 2 - 3 36 r + 1 35 C r

k 2 - 3 = r + 1 6 , k 2 - 3 > 0

(i) k = ± 2 gives r = 5

(ii) k = ± 3 gives r = 35

4 ordered pairs

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A
alok kumar singh

Contributor-Level 10

? P? =? P? ⇒ n!/ (n-r)! = n!/ (n-r-1)! ⇒ n-r=1.
? C? =? C? ⇒ r + (r-1) = n ⇒ n = 2r-1.
Substitute n: (2r-1)-r=1 ⇒ r=2.

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a month ago

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A
alok kumar singh

Contributor-Level 10

I (6)    F (8)

Case I       2            4

Case II      3            6

Case III     4            8

Total =   

 

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a month ago

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A
alok kumar singh

Contributor-Level 10

  r = 1 n t a n 1 1 1 + r + r 2 = r = 1 n t a n 1 ( r + 1 ) 1 + r ( r + 1 )

= t a n 1 2 t a n 1 + . . . . . . + t a n 1 ( n + 1 ) t a n 1 n            

For n  value = π 2 π 4 = π 4  

t a n ( π 4 ) = 1           

New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

A            B            C

-                -            1

-                -            2

-                -            3

Number of groups = 1 0 C 1 ( 2 9 2 ) = 5 1 0 0  

...more

New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

Number divisible by 3;

(a) sum of the digit must be divisible by 3

(i)  1 2 3  3! = 6

(ii) 1, 3, 5 -> 3! = 6

(iii) 2, 3, 4 -> 3! = 6

(iv) 3, 4, 5 = 3! = 6

Total = 24

(b) Divisible by

5  4 * 3 = 12

(c) Now common divisible by both

1 3 5 2! = 2

For 3, 4  5  2! = 2

Total ways = 24 + 12 – 4 = 32

 

New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

xyz = 24

24 = 23 * 3

Let's distribute 2, 3 among 3 variables. No. of positive integral solution =

No. of ways to distribute = 

 

          

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a month ago

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