Quantitative Aptitude — Permutation and Combination

The word 'APPLE' has 5 letters. The letter 'P' is repeated 2 times. The formula for permutations of n items where p1 items are of one type, p2 of another, etc., is n! / (p1! * p2! * ...). Here, n = 5 (total letters). p1 = 2 (for 'P'). Number of different words = 5! / 2! = (5 * 4 * 3 * 2 * 1) / (2 * 1) = 120 / 2 = 60.

This is a combination problem as the order of selection does not matter. The formula for combinations of selecting r items from n items is nCr = n! / (r! * (n-r)!). Here, n = 7 (total people), r = 3 (members to be selected). Number of ways = 7C3 = 7! / (3! * (7-3)!) = 7! / (3! * 4!) = (7 * 6 * 5 * 4 * 3 * 2 * 1) / ((3 * 2 * 1) * (4 * 3 * 2 * 1)) = (7 * 6 * 5) / (3 * 2 * 1) = 7 * 5 = 35.

This is a permutation problem as the order of digits matters (e.g., 123 is different from 321). We need to arrange 3 digits out of 5 distinct digits. The formula for permutations of selecting r items from n distinct items is nPr = n! / (n-r)!. Here, n = 5 (total digits), r = 3 (digits for the number). Number of 3-digit numbers = 5P3 = 5! / (5-3)! = 5! / 2! = (5 * 4 * 3 * 2 * 1) / (2 * 1) = 5 * 4 * 3 = 60.

The word 'EXAMINATION' has 11 letters. Vowels: E, A, I, A, I, O (6 vowels). Consonants: X, M, N, T, N (5 consonants). Repeated letters: 'A' appears 2 times, 'I' appears 2 times, 'N' appears 2 times. Step 1: Treat all vowels as a single block. The vowels are E, A, I, A, I, O. The number of ways to arrange these 6 vowels among themselves is 6! / (2! * 2!) (due to 'A' and 'I' repeating twice). Arrangement of vowels = (6 * 5 * 4 * 3 * 2 * 1) / ((2 * 1) * (2 * 1)) = 720 / 4 = 180 ways. Step 2: Arrange the vowel block and the consonants. The units to be arranged are (E A I A I O), X, M, N, T, N. This makes a total of 1 (vowel block) + 5 (consonants) = 6 units. Among these 6 units, the consonant 'N' is repeated 2 times. The number of ways to arrange these 6 units is 6! / 2! (due to 'N' repeating twice). Arrangement of units = (6 * 5 * 4 * 3 * 2 * 1) / (2 * 1) = 720 / 2 = 360 ways. Step 3: Total number of words. Multiply the arrangements of vowels within their block by the arrangements of the units. Total words = (Arrangement of vowels) * (Arrangement of units) = 180 * 360 = 64800.

Total red balls = 5. Total blue balls = 4. Total balls = 9. We need to select 3 balls such that at least 2 are red. This means we can have: Case 1: Exactly 2 red balls and 1 blue ball. Number of ways = (Number of ways to choose 2 red from 5) * (Number of ways to choose 1 blue from 4) = 5C2 * 4C1 = ((5 * 4) / (2 * 1)) * (4 / 1) = 10 * 4 = 40 ways. Case 2: Exactly 3 red balls and 0 blue balls. Number of ways = (Number of ways to choose 3 red from 5) * (Number of ways to choose 0 blue from 4) = 5C3 * 4C0 = ((5 * 4) / (2 * 1)) * 1 (Since 4C0 = 1) = 10 * 1 = 10 ways. Total number of ways = Ways in Case 1 + Ways in Case 2 = 40 + 10 = 50 ways.

Digits available: 0, 1, 2, 3, 4, 5 (6 digits). For a number to be even, its last digit must be 0, 2, or 4. Also, the first digit cannot be 0 for a 4-digit number. Case 1: The last digit is 0. _ _ _ 0 The last digit is fixed as 0 (1 way). Remaining 5 digits (1, 2, 3, 4, 5) are available for the first three places. Number of ways to fill the first 3 places = 5P3 = 5 * 4 * 3 = 60 ways. So, numbers ending with 0 = 60. Case 2: The last digit is 2 or 4. Subcase 2a: The last digit is 2. _ _ _ 2 The last digit is fixed as 2 (1 way). Now, we have 5 remaining digits (0, 1, 3, 4, 5). The first digit cannot be 0. So, the first digit can be chosen from {1, 3, 4, 5} (4 ways). After choosing the first digit, we have 4 digits remaining (including 0). The second digit can be chosen in 4 ways. The third digit can be chosen in 3 ways. So, numbers ending with 2 = 4 * 4 * 3 = 48 ways. Subcase 2b: The last digit is 4. _ _ _ 4 Similar to Subcase 2a, the last digit is fixed as 4 (1 way). The first digit can be chosen from {1, 2, 3, 5} (4 ways). The second digit can be chosen in 4 ways. The third digit can be chosen in 3 ways. So, numbers ending with 4 = 4 * 4 * 3 = 48 ways. Total number of even numbers = (Numbers ending with 0) + (Numbers ending with 2) + (Numbers ending with 4) = 60 + 48 + 48 = 156.

Total men = 6. Total women = 4. Committee size = 5 members. Condition: Exactly 2 women must be included. If there are exactly 2 women, then the remaining (5 - 2) = 3 members must be men. Step 1: Select 2 women from 4 women. Number of ways = 4C2 = 4! / (2! * (4-2)!) = 4! / (2! * 2!) = (4 * 3 * 2 * 1) / ((2 * 1) * (2 * 1)) = 12 / 2 = 6 ways. Step 2: Select 3 men from 6 men. Number of ways = 6C3 = 6! / (3! * (6-3)!) = 6! / (3! * 3!) = (6 * 5 * 4 * 3 * 2 * 1) / ((3 * 2 * 1) * (3 * 2 * 1)) = (6 * 5 * 4) / (3 * 2 * 1) = 20 ways. Step 3: Total ways to form the committee. Since both selections must happen, we multiply the number of ways. Total ways = (Ways to select women) * (Ways to select men) = 6 * 20 = 120 ways.

Condition: No two girls sit together. To ensure no two girls sit together, we first arrange the boys, and then place the girls in the spaces between the boys or at the ends. Step 1: Arrange the 5 boys. The 5 boys can be arranged in 5! ways. 5! = 5 * 4 * 3 * 2 * 1 = 120 ways. Step 2: Create spaces for the girls. When 5 boys are arranged in a row, they create 6 possible spaces where the girls can sit so that no two girls are together. _ B _ B _ B _ B _ B _ (The underscores represent the possible spaces for girls). Step 3: Arrange the 4 girls in these 6 spaces. We need to choose 4 spaces out of 6 available spaces and arrange the 4 girls in those chosen spaces. This is a permutation problem (order matters for girls). Number of ways = 6P4 = 6! / (6-4)! = 6! / 2! = (6 * 5 * 4 * 3 * 2 * 1) / (2 * 1) = 6 * 5 * 4 * 3 = 360 ways. Step 4: Total number of ways. Multiply the number of ways to arrange boys by the number of ways to arrange girls. Total ways = (Arrangement of boys) * (Arrangement of girls in spaces) = 120 * 360 = 43200 ways.

Total people = 7. Condition: Two particular people must always sit together. Step 1: Treat the two particular people as a single unit. Now, we have (7 - 2 + 1) = 6 units to arrange around the circular table. The formula for circular permutations of n distinct items is (n-1)!. So, the number of ways to arrange these 6 units circularly is (6 - 1)! = 5!. 5! = 5 * 4 * 3 * 2 * 1 = 120 ways. Step 2: Arrange the two particular people within their unit. The two particular people can arrange themselves within their unit in 2! ways. 2! = 2 * 1 = 2 ways. Step 3: Total number of ways. Multiply the arrangements of the units by the arrangements within the unit. Total ways = (Circular arrangement of units) * (Arrangement within the unit) = 120 * 2 = 240 ways.

Total men = 8. Total women = 7. Committee size = 6 members. Conditions: 1. At least 3 men (i.e., 3, 4, 5, or 6 men). 2. At least 2 women (i.e., 2, 3, 4, 5, or 6 women). Let's list the possible combinations of men and women that satisfy both conditions and sum up to 6 members: Case 1: 3 Men and 3 Women Number of ways to choose 3 men from 8 = 8C3 = 8! / (3! * 5!) = (8 * 7 * 6) / (3 * 2 * 1) = 56 ways. Number of ways to choose 3 women from 7 = 7C3 = 7! / (3! * 4!) = (7 * 6 * 5) / (3 * 2 * 1) = 35 ways. Ways for Case 1 = 56 * 35 = 1960. Case 2: 4 Men and 2 Women Number of ways to choose 4 men from 8 = 8C4 = 8! / (4! * 4!) = (8 * 7 * 6 * 5) / (4 * 3 * 2 * 1) = 70 ways. Number of ways to choose 2 women from 7 = 7C2 = 7! / (2! * 5!) = (7 * 6) / (2 * 1) = 21 ways. Ways for Case 2 = 70 * 21 = 1470. Other possibilities (not allowed): - (5 Men, 1 Woman) violates "at least 2 women". - (6 Men, 0 Women) violates "at least 2 women". - (2 Men, 4 Women) violates "at least 3 men". Total number of ways to form the committee = Ways for Case 1 + Ways for Case 2 = 1960 + 1470 = 3430 ways.