If nC6 ÷ n-2C3 = 91/4 then the value of n is _____________.
In forming a committee of 5 out of 5 males and 6 females how many choices you have to made so that there are 3 males and 2 females?
5C3 x 6C2 = 5 x 4/2 x 6 x 5/2
=150
In order to pass PE-II examination minimum marks have to be secured in each of 7 subjects. In how many ways can pupil fail?
Given that for a student to pass an examination , a minimum has to be secured in each of the 7 subjects.
A student would fail if he doesn't secure a minimum in any one o
the subjects A student would fail either if he fails in 1 subject or he fails
in 2 subjects or 3 subjects or so on continuing.........or fails in all the
subjects
A student can fail in 1 subject in ⁷C₁ ways
A student can fail in 2 subjects in ⁷C₂ ways
A student can fail in 3 subjects in ⁷C₃ ways
A student can fail in all subjects in ⁷C₇ = 1 way
So, total number of ways in which student would fail an examination are
(⁷C₁ + ⁷C₂ + .......+ ⁷C₇) = 2⁷ - 1 = 127
A team of 12 men is to be formed out of n persons. It is found that ‘A’ and ‘B’ are three times as often together as ‘C’ ‘D’ and ‘E’ are. Then the value of n is __________.
You are selecting a cricket team of first 11 players out of 16 including 4 bowlers and 2 wicket keepers. In how many ways you can do it so that the team contains exactly 3 bowlers and 1 wicket keeper?
Out of 10 consonants and 4 vowels how many words can be formed each containing 6 consonant and 3 vowels?
In your college Union Election you have to choose candidates. Out of 5 candidates 3 are to be elected and you are entitled to vote for any number of candidates but not exceeding the number to be elected. You can do it in ________ ways.
A team of 12 men is to be formed out of n persons. Then the number of times 2 men ‘A’ and ‘B’ are together is ________.
In paper from 2 groups of 5 questions each you have to answer any 6 questions attempting at least 2 question from each group. This is possible in ________ number of ways.
The number of combinations that can be made by taking 4 letters of the word 'COMBINATION'
Out of 8 different balls taken three at a time without taking the same three together more than once for how many number of times you can select a particular balls?
A boat’s crew consist of 8 men, 3 of whom can row only on one side and 2 only on the other. The number of ways in which the crew can be arranged is _______.
Out of 8 different balls taken three at a time without taking the same three together more than once for for how many number of times you can select any ball?
You are selecting a cricket team of first 11 players out of 16 including 4 bowlers and 2 wicket keepers. In how many ways you can do it so that the team contains exactly 3 bowlers and 1 wicket keeper? Would your answer be different if the team contains at least 3 bowlers and at least 1 wicket-keeper?
ANSWER :- A
Solution :-
There are : 4 bowlers, 2 wicket-keepers, 10 non-bowlers-not-wicket keepers.
There are exactly 3 bowlers, exactly 1 wicket keeper, and exactly 7 non-bowlers-non-wicket-keepers.
(4C3)(2C1)(10C7)
= 960
If 18Cn = 18Cn+2 then the value of n is _________.
In your office 4 posts have fallen vacant. In how many ways a selection out of 31 candidates can be made if one candidate is always included?
since 1 candidate is already selected so there are 3 posts for which we need candidates from 30.
A party of 6 is to be formed form 10 men an 7 women so as to include 3 men and 3 women. In how many ways the party can be formed if two particular women refuse to join it?
A team of 12 men is to be formed out of n person. The number of times 3 men ‘C’ ‘D’ and ‘E’ are together is_________.
Put C,D and E as one unit
∴ Total people to be selected = 10 = 9 ppl + one group of 3
Similarly total no ppl= n-3+1 = n-(C,D,E) + a group of 3 ppl [Note:Here 3 ppl are replaced by one group of 3 ppl ∴ there are counted as one and not 3]
∴
Selection:n-3+1 C10 = n-2C10
In how many ways can 8 boys form a ring?
The total number of sitting arrangements of 7 persons in a row if 3 persons sit together in a particular order is _________.
In how many ways 6 men can sit at a round table so that all shall not have the same neighbors in any two occasions?
From 6 boys and 4 girls 5 are to be seated. If there must be exactly 2 girls the number of ways of selection is __________.
If you have to make a choice of 7 questions out of 10 questions set, you can do it in ________ number of ways.
The total number of sitting arrangements of 7 persons in a row if one person occupies the middle seat is _________.
In how many ways 6 men and 6 women sit at a round table so that no two men are together?
First, we sit the men in an order that there is a place vacant between two men. Then, number of ways sitting 6 men = (6 – 1)! = 5!
Now a number of ways sitting 6 women in the vacant places = 6!
Number of permutations = (6 – 1)! = 5!
Now, the number of ways sitting 6 women in the vacant place = 6!
Number of permutaions = 5! × 6!
If all the permutations of the letters of the word “chalk” are written in a dictionary the rank of this word will be __________.
In how many ways 4 men and 3 women are arranged at a round table if the women always sit tohether?
If all three women always sit together
We can club the 3 women in one group and the number of arrangements of this group will be 3!
Now, 4 men and 1 group of women can be arranged arround a round table =(5−1)!
Hence, Total arrangements =4!3!=144.
The total number of sitting arrangements of 7 persons in a row if 2 persons occupy the end seats is _________.
A family comprised of an old man, 6 adults and 4 children is to be seated is a row with the condition that the children would occupy both the ends and never occupy either side of the old man. How many sitting arrangements are possible?
There are 11 persons and 11 seats.
Select 2 children(4C2 ways).
These two children can be seated in the side seats in 2! ways.
The old man can occupy any of the 7 middle seats(7 ways).
(note that he cannot occupy 2nd and 10th seat because these seats are near to the children sitting in the side seats)
3 seats are occupied and 8 seats are remaining. In these 8 seats, 2 seats will be near to the old man in which children cannot seat. Thus 6 seats are remaining in which 2 children can be seated. This can be done in 6P2 ways.
5 persons are seated now. Remaining 6 persons can be arranged in the remaining 6 seats in 6! ways.
Therefore, required number of ways
= 4C2 × 2! × 7 × 6P2 × 6!
= 6 × 2 × 7 × 30 × 720
= 1814400
In your office 4 posts have fallen vacant. In how many ways a selection out of 31 candidates can be made if one candidate is always included?
In a ration shop queue 2 boys, 2 girls, and 2 men are standing in such a way that the boys the girls and the men are together each.The total number of ways of arranging the queue is _______.
In how many ways 4 men and 3 women are arranged at a round table if the women always sit together?
For the given case : First fix the women in one group. 3 women can be seated in 3! ways. The 4 men can be seated in the remaining 4 places in 4! ways. Therefore, there are 144 arrangements.
Take women as 1 group. So we now have 4 men and 1 group . These 5 men can arranged in round table in (5-1)! = 4! = 24 ways.
But women can be further arranged in group in 3!= 6 ways.
Therefore total number of ways is equal to 24 × 6 = 144 ways .
Case 2- No women sit together.
As number of women is less than number of men , two of the men will be in between a pair of women. So fix the seats of women.
Women can be arranged in 3!= 6 ways.
Now select two men. They can be selected in C(4,2) = 6 ways and arranged in 2! = 2 ways.
Consider these 2 men as a single man. Now we have 3 men and they can be arranged in the round table in (3-1)! = 2! = 2 ways .
Therefore total number of ways = 6 × 6 × 2 × 2 = 144 ways .
The total number of sitting arrangements of 7 persons in a row if 3 persons sit together in any order is _________.
In your office 4 posts have fallen vacant. In how many ways a selection out of 31 candidates can be made if one candidate is always included?
In how many ways can 8 beads of different colour be strung on a ring?
How many numbers greater than a million can be formed with the digits: One 0 Two 1 one 3 and Three 7?
In how many ways the letters of the word "LOCKDOWN" can be arranged such that N is always between O's?
How many different words can be formed with the letter of the word “HARIYANA”?
How many number greater than 23000 can be formed with 1, 2, ………5?
How many different words can be formed with the letter of the word “HARIYANA”?How many arrangements are possible keeping 'h' and 'n' together ?
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