The total number of numbers, lying between 100 and 1000 that can be formed with the digits 1, 2, 3, 4, 5, if the repetition of digits is not allowed and numbers are divisible by either 3 or 5, is……………..
The total number of numbers, lying between 100 and 1000 that can be formed with the digits 1, 2, 3, 4, 5, if the repetition of digits is not allowed and numbers are divisible by either 3 or 5, is……………..
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1 Answer
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Number divisible by 3;
(a) sum of the digit must be divisible by 3
(i) 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
4 × 3 = 12
(c) Now common divisible by both
2! = 2
For 3, 4 2! = 2
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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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