There are 5 students in class 10, 6 students in class 11 and 8 students in class 12. If the number of ways, in which 10 students can be selected from them so as to include at least 2 students from each class and at most 5 students from the total 11 students of class 10 and 11 is 100k, then k is equal to…….

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  • R

    Answered by

    Raj Pandey | Contributor-Level 9

    a month ago

    | Class | No. of students | Number of possible cases |
    | :-: | :-: | :- |
    | 10 | 5 | (I) 2 | (II) 3 | (III) 2 |
    | 11 | 6 | (I) 2 | (II) 2 | (III) 3 |
    | 12 | 8 | (I) 6 | (II) 5 | (III) 5 |
    Total cases =? C? ×? C? ×? C? +? C? ×? C? ×? C? +? C? ×? C? ×? C?
    = 23,800 = 100K
    ∴ K = 238

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A
alok kumar singh

Start with

(1) E ¯ : 6 ! 2 ! = 3 6 0  

(2)    G E ¯ : 5 ! 2 ! , G N ¯ : 5 ! 2 !  

(3) GTE : 4!, GTN: 4!, GTT : 4!

(4) GTWENTY = 1

360 + 60 + 60 + 24 + 24 + 24 + 1 = 553

A
alok kumar singh

x + 2y + 3z = 42

0    x + 2y = 42 ->22 cases

1    x + 2y = 39 ->19 cases

2    x + 2y = 36 ->19 cases

3    x + 2y = 33 ->17 cases

4    x + 2y = 30 ->16 cases

5    x + 2y = 27 ->14 cases

6    x + 2y = 24 ->13 cases

7    x + 2y = 21 ->11 cases

8    x + 2y = 18 ->10 cases

9    x + 2y = 15 ->8 cases

10  x + 2y =12 -> 7 cases

11  x + 2y = 9 -> 5 cases

12  x + 2y = 6 -> 4 cases

13  x + 2y = 3 -> 2 cases

14  x + 2y = 0 -> 1 cases.

A
alok kumar singh

Total ways to partition 5 into 4 parts are:

5 0 

4 1 0  5!4!=5

3 2 0 5 ! 3 ! 2 ! = 1 0

3 1 0 5 ! 2 ! 2 ! 2 ! = 1 5

2 1 5 ! 2 ! × 3 ! = 1 0

51 Total way

A
alok kumar singh

After giving 2 apples to each child 15 apples left now 15 apples can be distributed in
15+3–1C2 = 17C2 ways

  = 1 7 * 1 6 2 = 1 3 6          

A
alok kumar singh

( n ? 1 ) ( n ? 2 ) ( n ? 3 ) n ( n ? 1 ) ( n ? 2 ) ( n ? 3 ) = 1 8
  n = 8

= 8 * 7 * 6 * 5 * 4 + 9 * 8 2

= 6756

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