mathemerize

Solution : Total Number of ways = \(^{10}C_1\) + \(^{10}C_2\) + \(^{10}C_3\) + \(^{10}C_4\) = 10 + 45 + 120 + 210 = 385 Similar Questions A student is to answer 10 out of 13 questions in an examination such that he must choose atleast 4 from the first five questions. The number of choices …

At an election, a voter may vote for any number of candidates not greater than the number of candidates to be elected. There are 10 candidates and 4 are to be elected. If a voter votes for atleast 1 candidate, then the number of ways in which he can vote is Read More »

Solution : first we choose 4 numbers from 12 numbers, then 4 from remaining 8 numbers, and then 4 from remaining 4 numbers So, Required number of ways = \(^{12}C_4\) x \(^8C_4\) x \(^4C_4\) = \(12!\over (4!)^3\) Similar Questions How many different words can be formed by jumbling the letters in the word ‘MISSISSIPPI’ in …

The set S = {1,2,3,…..,12} is to be partitioned into three sets A, B and C of equal size. Thus, \(A\cup B\cup C\) = S \(A\cap B\) = \(B\cap C\) = \(A\cap C\) = \(\phi\) The number of ways to partition S is Read More »

Solution : Given word is MISSISSIPPI, Here, I occurs 4 times, S = 4 times P = 2 times, M = 1 time So, we write it like this _M_I_I_I_I_P_P_ Now, we see that spaces are the places for letter S, because no two S can be together So, we can place 4 S in …

How many different words can be formed by jumbling the letters in the word ‘MISSISSIPPI’ in which no two S are adjacent ? Read More »

Solution : The number of ways in which 4 novels can be selected = \(^6C_4\) = 15 The number of ways in which 1 dictionary can be selected = \(^3C_1\) = 3 Now, we have 5 places in which middle place is fixed. \(\therefore\)  4 novels can be arranged in 4! ways \(\therefore\)  total number …

From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on the shelf so that the dictionary is always in the middle. Then, the number of such arrangements is Read More »

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 …

There are 10 points in a plane, out of these 6 are collinear. If N is the number of triangles formed by joining these points, then Read More »

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 …

Let \(T_n\) be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If \(T_{n+1}\) – \(T_n\) = 10, then the value of n is Read More »