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…….
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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1 Answer
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| 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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Start with
(1)
(2)
(3) GTE : 4!, GTN: 4!, GTT : 4!
(4) GTWENTY = 1
⇒ 360 + 60 + 60 + 24 + 24 + 24 + 1 = 553
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.
Total ways to partition 5 into 4 parts are:
5 0
4 1 0
3 2 0
3 1 0
2 1
51 Total way
After giving 2 apples to each child 15 apples left now 15 apples can be distributed in
15+3–1C2 = 17C2 ways
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