Solution : Given P(A \(\cup\) B)’ = 1/6, P(A \(\cap\) B) = 1/4 and P(A)’ = 1/4 \(\therefore\) P(A \(\cup\) B) = 1 – P(A \(\cup\) B)’ = 1 – \(1\over 6\) = \(5\over 6\) and P(A) = 1 – P(A)’ = 1 – \(1\over 4\) = \(3\over 4\) P(A \(\cup\) B) = P(A) + […]
Solution : The number of ways to n people on circular table is (n-1)! So, first we fix position of men, the number of ways to sit men = 5! Now, women can sit in the gaps between men, there are 6 gaps between 5 mens, So, women can sit in \(^6P_5\) ways Hence, total
Solution : The number of choices available to him = \(^5C_4\) x \(^8C_6\) + \(^5C_5\) x\(^8C_5\) = 5.4.7 + 8.7 = 140 + 56 = 196 Similar Questions The number of ways in which 6 men and 5 women can dine at a round table, if no two women are to sit together, is given
Solution : The number of ways of distributing n identical things in m different boxes is \(^{n-1}C_{m-1}\) So, Required number of ways = \(^{8-1}C_{3-1}\) = \(^7C_2\) = 21 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.
Solution : Total number of ways in which all the letters can be arranged in alphabetical order = 6! There are two vowels (A, E) in the word ‘GARDEN’. Total number of ways in which these two vowels can be arranged = 2! \(\therefore\) Total number of required ways = \(6!\over 2!\) = 360 Similar
Solution : In the word SACHIN’ order of alphabets is A, C, H, I, N and S. Number of words start with A = 5!, so with C, H, I, N. Now, words start with S and after that ACHIN are in ascending order of position, so 5.5! = 600 words are in dictionary before,
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
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
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
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
