Home > Practice > Arithmetic Aptitude > Permutations and Combinations > Miscellaneous
61. How many arrangements of four 0's (zeroes), two 1's and two 2's are there in which the first 1 occur before the first 2?
Since, there are two 1's and two 0's, the number of arrangements in which the first 1 is before the first 2 is same as the number of arrangement in which the first 2 is before the first 1 and they are each equal to half the total number of arrangements = 210
Solution:
Total number of arrangements = = 420Since, there are two 1's and two 0's, the number of arrangements in which the first 1 is before the first 2 is same as the number of arrangement in which the first 2 is before the first 1 and they are each equal to half the total number of arrangements = 210
62. How many different five-letter words can be formed using the letter from the world APPLE?
For example let us consider one permutation: P1LEAP2
Now if we permute the P’s amongst them we still get the same word PLEAP. The two P’s can be permuted amongst them in 2! ways.
We were counting P1LEAP2 and P2LEAP1 as different arrangements only because we were artificially distinguishing between the two P’s Hence the number of different five letter words that can be formed is:
=
= 60
Solution:
If the two P’s were distinct (they could have different subscripts and colours), the number of possible permutations would have been 5! = 120For example let us consider one permutation: P1LEAP2
Now if we permute the P’s amongst them we still get the same word PLEAP. The two P’s can be permuted amongst them in 2! ways.
We were counting P1LEAP2 and P2LEAP1 as different arrangements only because we were artificially distinguishing between the two P’s Hence the number of different five letter words that can be formed is:
=
= 60
63. I have an amount of Rs. 10 lakh, which I went to invest in stocks of some companies. I always invest only amounts that are multiples of Rs 1 lakh in the stock of any company. If I can choose from among the stocks of five different companies, In how many ways can I invest the entire amount that I have?
This case can be represented as arranging ten balls and (5 - 1) four walls in the single row, which can be done in 14C4 ways. (The balls placed between every successive pair of walls belong to one group) 14C4 = 1001 ways.
Solution:
The situation is similar to placing 10 identical balls among 5 distinguishable boxes, where a box may have zero or more balls in it.This case can be represented as arranging ten balls and (5 - 1) four walls in the single row, which can be done in 14C4 ways. (The balls placed between every successive pair of walls belong to one group) 14C4 = 1001 ways.
64. A selection committee is to be chosen consisting of 5 ex-technicians. Now there are 12 representatives from four zones. It has further been decided that if Mr. X is selected, Y and Z will not be selected and vice-versa. In how many ways it can be done?
10C4 : when one of them is included.
Number of ways
= 10C5 + 10C4 + 10C4
= 672
Solution:
10C5 : when both are not included.10C4 : when one of them is included.
Number of ways
= 10C5 + 10C4 + 10C4
= 672
65. How many 5-digit positive integers exist the sum of whose digits are odd?
Out of these 90000 positive integers, the sum of the digits of half of the numbers will add up to an odd number and the remaining half will add up to an even number.
Hence, there are = 45000, 5-digit positive integers whose sum add up to an odd number.
Solution:
There are 9 × 104 = 90000, 5-digit positive integers.Out of these 90000 positive integers, the sum of the digits of half of the numbers will add up to an odd number and the remaining half will add up to an even number.
Hence, there are = 45000, 5-digit positive integers whose sum add up to an odd number.
66. There are eight boxes of chocolates, each box containing distinct number of chocolates from 1 to 8. In how many ways four of these boxes can be given to four persons (one boxes to each) such that the first person gets more chocolates than each of the three, the second person gets more chocolates than the third as well as the fourth persons and the third person gets more chocolates than fourth person?
For each combination of 4 out of 8 boxes, the box with the greatest number has to be given to the first person, the box with the second highest to the second person and so on.
The number of ways of giving 4 boxes to the 4 person is,
8C4 = 70
Solution:
All the boxes contain distinct number of chocolates.For each combination of 4 out of 8 boxes, the box with the greatest number has to be given to the first person, the box with the second highest to the second person and so on.
The number of ways of giving 4 boxes to the 4 person is,
8C4 = 70
67. In how many ways can seven friends be seated in a row having 35 seats, such that no two friends occupy adjacent seats?
They create 29 slots - one on the left of each seat and one on the right of the last one.
We can place the 7 friends in any of these 29 slots i.e. 29P7 ways.
Solution:
First let us consider the 28 unoccupied seats.They create 29 slots - one on the left of each seat and one on the right of the last one.
We can place the 7 friends in any of these 29 slots i.e. 29P7 ways.
68. How many ways can 10 letters be posted in 5 post boxes, if each of the post boxes can take more than 10 letters?
So, the first letter has 5 options, so does the second letter and so on and so forth for all of the 10 letters.
i.e.5 × 5 × 5 × . . . . × 5 (up-to 10 times).
= 510
Solution:
Each of the 10 letters can be posted in any of the 5 boxes.So, the first letter has 5 options, so does the second letter and so on and so forth for all of the 10 letters.
i.e.5 × 5 × 5 × . . . . × 5 (up-to 10 times).
= 510
69. In a hockey championship, there are 153 matches played. Every two team played one match with each other. The number of teams participating in the championship is:
nC2 = 153
On solving we get,
⇒ n = −17 and n =18
It cannot be negative so,
n = 18 is the answer.
Solution:
Let there were x teams participating in the games, then total number of matches,nC2 = 153
On solving we get,
⇒ n = −17 and n =18
It cannot be negative so,
n = 18 is the answer.
70. A box contains 10 balls out of which 3 are red and rest are blue. In how many ways can a random sample of 6 balls be drawn from the bag so that at the most 2 red balls are included in the sample and no sample has all the 6 balls of the same colour?
One red ball and 5 blue ball
Or
Two red balls and 4 blue balls.
Total number of ways:
= (3C1 × 7C5) + (3C2 × 7C4)
= 63 + 105
= 168
Solution:
Six balls can be selected in the following ways:One red ball and 5 blue ball
Or
Two red balls and 4 blue balls.
Total number of ways:
= (3C1 × 7C5) + (3C2 × 7C4)
= 63 + 105
= 168