Solution : If out of n points, m are collinear, then Number of triangles = \(^nC_3\) – \(^mC_3\) \(\therefore\) Number of triangles = \(^{10}C_3\) – \(^6C_3\) = 120 – 20 = 100 Similar Questions How many different words can be formed by jumbling the letters in the word ‘MISSISSIPPI’ in which no two S are …
Solution : Given, \(T_n\) = \(^nC_3\) \(T_{n+1}\) = \(^{n+1}C_3\) \(\therefore\) \(T_{n+1}\) – \(T_n\) = \(^{n+1}C_3\) – \(^{n}C_3\) = 10 [given] \(\therefore\) \(^nC_2\) + \(^nC_3\) – \(^nC_3\) = 10 \(\implies\) \(^nC_2\) = 10 \(\therefore\) n = 5 Similar Questions How many different words can be formed by jumbling the letters in the word ‘MISSISSIPPI’ in which …
Solution : n(S) = \(^{10}C_4\) = 210 n(E)= \(^5C_2 \times ^3C_1 \times ^2C_1\) + \(^5C_1 \times ^3C_2 \times ^2C_1\) + \(^5C_1 \times ^3C_1 \times ^2C_2\) = 105 \(\therefore\) P(E) = \(105\over 210\) = \(1\over 2\) Similar Questions How many different words can be formed by jumbling the letters in the word ‘MISSISSIPPI’ in which no …
Solution : 5 different mangoes can be distributed by following ways among 3 children such that each gets at least 1 : Total number of ways : (\(5!\over 3!1!1!2!\) + \(5!\over 2!2!2!\)) \(\times\) 3! Now, the number of ways of distributing remaining fruits (i.e. 4 oranges + 3 apples) among 3 children = \(3^7\) (as …
Solution : There are 4 odd digits (1, 1, 3, 3) and 4 odd places(first, third, fifth and seventh). At these places the odd digits can be arranged in \(4!\over 2!2!\) = 6 Then at the remaining 3 places, the remaining three digits(2, 2, 4) can be arranged in \(3!\over 2!\) = 3 ways Therefore, …
Solution : First of all, arrange all letters of given word alphabetically : ‘ADIPR’ Total number of words starting with A _ _ _ _ = 4! = 24 Total number of words starting with D _ _ _ _ = 4! = 24 Total number of words starting with I _ _ _ _ …
