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S seema garhwal
In the word ASSASSINATION, we have number of S =4 number of A =3 number of I= 2 number of N =2  Rest of letters appear at once. Since all words have to be arranged in such a way that all the S are together so we can assume SSSS as an object. The single object SSSS with other 9 objects is counted as 10. These 10 objects can be arranged in  (we have 3 A's,2 I's,2 N's)                            ...

S seema garhwal
From a class of 25 students, 10 are to be chosen for an excursion party. There are 3 students who decide that either all of them will join or none of them will join, there are two cases : The case I: All 3 off them join. Then, the remaining 7 students can be chosen from 22 students in  ways. Case II : All 3 of them do not join. Then,10 students can be chosen from 22 students in   ways. Thus,...

S seema garhwal
It is required to seat 5 men and 4 women in a row so that the women occupy the even places. The 5 men can be seated in  ways. 4 women can be seated at cross marked places (so that women occupy even places) Therefore, women can be seated in  ways. Thus, the possible arrangements

S seema garhwal
From a deck of 52 cards, 5 cards combinations have to be made in such a way that in each selection of 5 cards there is exactly 1 king. Number of kings =4  Number of ways of selecting 1 king  4 cards from the remaining 48 cards are selected in  ways. Thus, the required number of 5 card combinations

S seema garhwal
It is given that a question paper consists of 12 questions divided in two parts i.e., Part I and Part II, containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. This can be done as follows: (i) 3 questions from part I and 5 questions from part II (ii) 4 questions from part I and 4 questions from part II (iii)...

S seema garhwal
Two different vowels and 2 different consonants are to be selected from the English alphabets. Since there are 5 different vowels so the number of ways of selecting two different vowels =                                                                                                                                         Since there are 21 different consonants so the number of ways of...

S seema garhwal
For the number to be divisible by 10, unit digit should be 0. Thus, 0 is fixed at a unit place. Therefore, the remaining 5 places should be filled with 1,3,5,7,9. The remaining 5 vacant places can be filled in  ways. Hence, the required number of 6 digit numbers which are divisible by 10

S seema garhwal
In the word EXAMINATION, we have 11 letters out of which A,I, N  appear twice and all other letters appear once. The word that will be listed before the first word starting with E will be words starting with A. Therefore, to get the number of words starting with A, letter A is fixed at extreme left position, the remaining 10 letters can be arranged. Since there are 2 I's and 2 N's in the...

S seema garhwal
There are 9 boys and 4 girls. A committee of 7 has to be formed. (ii) atmost 3 girls, there can be 4 cases : (a) Girls =0, so boys in committee= 7-0=7 Thus, the required number of ways                                                                                                         (b) Girls =1, so boys in committee= 7-1=6 Thus, the required number of ways                                 ...

S seema garhwal
There are 9 boys and 4 girls. A committee of 7 has to be formed. (ii) at least 3 girls, there can be two cases : (a) Girls =3, so boys in committee= 7-3=4 Thus, the required number of ways                                                                                                                                                       (b) Girls =4, so boys in committee= 7-4=3 Thus, the...

S seema garhwal
There are 9 boys and 4 girls. A committee of 7 has to be formed. Given : Girls =3, so boys in committee= 7-3=4 Thus, the required number of ways

S seema garhwal
In the word EQUATION, we have  vowels = 5(A,E,I,O,U) consonants = 3(Q,T,N) Since all the vowels and consonants occur together so (AEIOU) and (QTN) can be assumed as single objects. Then, permutations of these two objects taken at a time  Corresponding to each of these permutations, there are  permutations for vowels and  permutations for consonants. Thus, by multiplication principle, required...

S seema garhwal
In the word DAUGHTER, we have  vowels = 3(A,E,U) consonants = 5(D,G,H,T,R) Number of ways of selecting 2 vowels  Number of ways of selecting 3 consonants  Therefore, the number of ways of selecting 2 vowels and 3 consonants                                                                                                              Each of these 30 combinations of 2 vowels and 3 consonants can...

S seema garhwal
9 courses are available and 2 specific courses are compulsory for every student. Therefore, every student has to select 3 courses out of the remaining 7 courses. This can be selected in  ways. Thus, using multiplication priciple, number of ways of selecting courses                                                                                                                                   ...

S seema garhwal
A bag contains 5 black and 6 red balls. 2 black balls can be selected in   ways and 3 red balls can be selected in  ways. Thus, using multiplication priciple, number of ways of selecting 2 black and 3 red balls                                                                                                                                                                                         ...

S seema garhwal
Out off, 17 players, 5 are bowlers. A cricket team of 11 is to be selected such that there are exactly 4 bowlers. 4 bowlers can be selected in   ways and 7 players can be selected in  ways. Thus, using multiplication priciple, number of ways of selecting the team                                                                                                                                      ...

S seema garhwal
In a deck, there is 4 ace out of 52 cards. A combination of 5 cards is to be selected containing exactly one ace. Then, one ace can be selected in  ways and other 4 cards can be selected in  ways. Hence, using the multiplication principle, required the number of 5 card combination                                                                                                                   ...

S seema garhwal
There are 6 red balls, 5 white balls and 5 blue balls. 9 balls have to be selected in such a way that consists of 3 balls of each colour. 3 balls are selected from 6 red balls in . 3 balls are selected from 5 white balls in  3 balls are selected from 5 blue balls in . Hence, by the multiplication principle, the number of ways of selecting 9 balls                                                 ...