Permutations and Combinations Level - 2
Q1–10 of 50How many even numbers of four digits can be formed with the digits 1, 2, 3, 4, 5, 6 (repetitions of digits are allowed)?
1. Number of even numbers = 6 × 6 × 6 × 3
How many 4 digit numbers divisible by 5 can be formed with the digits 0, 1, 2, 3, 4, 5, 6 and 6?
2. We need to think of this as: Number with two sixes or numbers with one six or number with no six.
0, 1, 2, 3, 4, 5, 6 and 6
Numbers with 2 sixes:
Numbers ending in zero \({}^{5}C_{1}\) × 3!/2! = 15
Numbers Ending in 5 and
(a) Starting with 6 \({}^{5}C_{1}\) × 2! = 10
(b) Not starting with 6 \({}^{4}C_{1}\) (as zero is not allowed) = 4
Number with 1 six or no sixes.
Numbers ending in 0 \({}^{6}C_{3}\) × 3! = 120
Numbers ending in 5 \({}^{5}C_{1}\) × \({}^{5}C_{2}\) × 2! = 100
Thus a total of 249 numbers.
There are 6 pups and 4 cats. In how many ways can they be seated in a row so that no cats sit together?
3. First arrange 6 pups in 6 places in 6! ways.
This will leave us with 7 places for 4 cats. Answer = 6! × \({}^{7}P_{4}\).
How many new words can be formed with the word MANAGEMENT all ending in G?
4. Arrangement of M, A, N, A, E, M, E, N, T is \(\frac{9!}{2!\times 2!\times 2!\times 2!}\).
Find the total numbers of 9-digit numbers that can be formed all having different digits.
5. For nine places we have following number of arrangements.
9 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2
There are $V$ lines parallel to the $x$-axis and '$W$' lines parallel to $y$-axis. How many rectangles can be formed with the intersection of these lines?
6. For a rectangle, we need two pair of parallel lines which are perpendicular to each other. We need to select two parallel lines from 'v' lines and 2 parallel lines from 'w' lines. Hence required number of parallel lines is \({}^{v}C_{2}\) × \({}^{w}C_{2}\).
From 4 gentlemen and 4 ladies a committee of 5 is to be formed. Find the number of ways of doing so if the committee consists of a president, a vice-president and three secretaries?
7. From 8 people we have to arrange a group of 5 in which three are similar \(\frac{{}^8P_5}{3!}\) or \(\frac{{}^8C_5\times 5!}{3!}\).
In the above question, what will be the number of ways of selecting the committee with at least 3 women such that at least one woman holds the post of either a president or a vice-president?
8. \(\frac{4C_4\times 4C_1\times 5!}{3!} + \frac{4C_2\times 4C_3\times 5!}{3!} - 4C_3\times 4C_2\times 2C_2\times 2!\)
Find the number of ways of selecting the committee with a maximum of 2 women and having at the maximum one woman holding one of the two posts on the committee.
9. Since the number of men and women in the question is the same, there is no difference in solving this question and solving the previous one (question number 8) as committees having a maximum of 2 women would mean committees having a minimum of 3 men and committees having at maximum one woman holding the post of either president or vice president would mean at least 1 man holding one of the two posts.
Thus, the answer would be:
Number of committees with 4 men and 1 woman (including all arrangements of the committees) + Number of committees with 3 men and 2 women (including all arrangements of the committees)–Number of committees with 3 men and 2 women where both the women are occupying the two posts.
= \((^{4}C_{4}\) × \({}^{4}C_{1}\) × 5!)/3! + \((^{4}C_{3}\) × \({}^{4}C_{2}\) × 5!)/3! – \((^{4}C_{3}\) × \({}^{4}C_{2}\) × \({}^{2}C_{2}\) × 2!) = 80 + 480 – 48 = 512
The crew of an 8 member rowing team is to be chosen from 12 men, of which 3 must row on one side only and 2 must row on the other side only. Find the number of ways of arranging the crew with 4 members on each side.
10. \({}^{7}C_{1}\) × \({}^{6}C_{2}\) × 4! × 4! = 60480