Permutations and Combinations
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New answer posted
a month agoContributor-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 agoContributor-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
New answer posted
a month agoContributor-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.
New answer posted
a month agoContributor-Level 10
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
New answer posted
a month agoContributor-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 agoTaking an Exam? Selecting a College?
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