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- Question
A box contains 4 different black balls, 3 different red balls and 5 different blue balls. In how many ways can the balls be selected if every selection must have at least 1 black ball and one red ball ? A) 24 - 1 B) 2425-1 C) (24-1)(23-1)25 D) None
Options- A. A
- B. B
- C. C
- D. D
- Correct Answer
- C
ExplanationIt is explicitly given that all the 4 black balls are different, all the 3 red balls are different and all the 5 blue balls are different. Hence this is a case where all are distinct objects.
Initially let's find out the number of ways in which we can select the black balls. Note that at least 1 black ball must be included in each selection.
Hence, we can select 1 black ball from 4 black balls
or 2 black balls from 4 black balls.
or 3 black balls from 4 black balls.
or 4 black balls from 4 black balls.Hence, number of ways in which we can select the black balls
= 4C1 + 4C2 + 4C3 + 4C4
= ........(A)Now let's find out the number of ways in which we can select the red balls. Note that at least 1 red ball must be included in each selection.
Hence, we can select 1 red ball from 3 red balls
or 2 red balls from 3 red balls
or 3 red balls from 3 red ballsHence, number of ways in which we can select the red balls
= 3C1 + 3C2 + 3C3
= ........(B)Hence, we can select 0 blue ball from 5 blue balls (i.e, do not select any blue ball. In this case, only black and red balls will be there)
or 1 blue ball from 5 blue balls
or 2 blue balls from 5 blue balls
or 3 blue balls from 5 blue balls
or 4 blue balls from 5 blue balls
or 5 blue balls from 5 blue balls.Hence, number of ways in which we can select the blue balls
= 5C0 + 5C1 + 5C2 + ? + 5C5
= ..............(C)From (A), (B) and (C), required number of ways
= - 1. Find the sum of the all the numbers formed by the digits 2,4,6 and 8 without repetition. Number may be of any of the form like 2,24,684,4862 ?
Options- A. 133345
- B. 147320
- C. 13320
- D. 145874 Discuss
- 2. There are 3 bags, in 1st there are 9 Mangoes, in 2nd 8 apples & in 3rd 6 bananas. There are how many ways you can buy one fruit if all the mangoes are identical, all the apples are identical, & also all the Bananas are identical ?
Options- A. 23
- B. 432
- C. 22
- D. 431 Discuss
- 3. What is the sum of all 3 digits number that can be formed using digits 0,1,2,3,4,5 with no repitition ?
Options- A. 28450
- B. 26340
- C. 32640
- D. 36450 Discuss
- 4. In how many ways can 100 soldiers be divided into 4 squads of 10, 20, 30, 40 respectively?
Options- A. 1700
- B. 18!
- C. 190
- D. None of these Discuss
- 5. Jay wants to buy a total of 100 plants using exactly a sum of Rs 1000. He can buy Rose plants at Rs 20 per plant or marigold or Sun flower plants at Rs 5 and Re 1 per plant respectively. If he has to buy at least one of each plant and cannot buy any other type of plants, then in how many distinct ways can Jay make his purchase?
Options- A. 3
- B. 6
- C. 4
- D. 2 Discuss
- 6. A group of 10 representatives is to be selected out of 12 seniors and 10 juniors. In how many different ways can the group be selected if it should have at least one senior ?
Options- A. ²²C?? + 1
- B. ²²C? + ¹?C?
- C. ²²C??
- D. ²²C?? - 1 Discuss
- 7. The number of permutations of the letters of the word 'MESMERISE' is ?
Options- A. 9!/(2!)^{2}x3!
- B. 9! x 2! x 3!
- C. 0
- D. None Discuss
- 8. A group consists of 4 men, 6 women and 5 children. In how many ways can 2 men , 3 women and 1 child selected from the given group ?
Options- A. 600
- B. 610
- C. 609
- D. 599 Discuss
- 9. In how many ways word of 'GLACIOUS' can be arranged such that 'C' always comes at end?
Options- A. 3360
- B. 5040
- C. 720
- D. 1080 Discuss
- 10. How many 4-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?
Options- A. 60
- B. 48
- C. 36
- D. 20 Discuss
Permutation and Combination problems
Search Results
Correct Answer: 147320
Explanation:
Sum of 4 digit numbers = (2+4+6+8) x x (1111) = 20 x 6 x 1111 = 133320
Sum of 3 digit numbers = (2+4+6+8) x
x (111) = 20 x 6 x 111 = 13320
Sum of 2 digit numbers = (2+4+6+8) x
x (11) = 20 x 3 x 11 = 660
Sum of 1 digit numbers = (2+4+6+8) x
x (1) = 20 x 1 x 1 = 20
Adding All , Sum = 147320
Correct Answer: 23
Explanation:
As in this problem , buying any fruit is different case , as buying apple is independent from buying banana. so ADDITION rule will be used.
= 23 will be answer.
Correct Answer: 32640
Explanation:
We know that zero can't be in hundreds place. But let's assume that our number could start with zero.
The formula to find sum of all numbers in a permutation is
111 x no of ways numbers can be formed for a number at given position x sum of all given digits
No of 1 s depends on number of digits
So,the answer us
111 x 20 x (0+1+2+3+4+5) = 33300
We got 20 as follows. If we have 0 in units place we can form a number in 4*5 ways. This is for all numbers. So we have substituted 20 in formula.
Now, this is not the final answer because we have included 0 in hundreds place. so we have to remove the sum of all numbers that starts with 0.
This is nothing but the sum of all 2 digits numbers formed by 1 2 3 4 5. Because 0 at first place makes it a 2 digit number.
So the sum for this is 11 x 4 x (1+2+3+4+5).
=660
Hope u understood why we use 4. Each number can be formed in 4x1 ways
So, the final answer is 33300-660 = 32640
Correct Answer: None of these
Explanation:
Correct Answer: 3
Explanation:
Let the number of Rose plants be ?a?.
Let number of marigold plants be ?b?.
Let the number of Sunflower plants be ?c?.
20a+5b+1c=1000; a+b+c=100
Solving the above two equations by eliminating c,
19a+4b=900
b = (900-19a)/4
b = 225 - 19a/4----------(1)
b being the number of plants, is a positive integer, and is less than 99, as each of the other two types have at least one plant in the combination i.e .:0 < b < 99--------(2)
Substituting (1) in (2),
0 < 225 - 19a/4 < 99
225 < -19a/4 < (99 -225)
=> 4 x 225 > 19a > 126 x 4
=> 900/19 > a > 505
a is the integer between 47 and 27 ----------(3)
From (1), it is clear, a should be multiple of 4.
Hence possible values of a are (28,32,36,40,44)
For a=28 and 32, a+b>100
For all other values of a, we get the desired solution:
a=36,b=54,c=10
a=40,b=35,c=25
a=44,b=16,c=40
Three solutions are possible.
Correct Answer: ²²C?? - 1
Explanation:
The total number of ways of forming the group of ten representatives is ²²C??.
The total number of ways of forming the group that consists of no seniors is ¹?C?? = 1 way
The required number of ways = ²²C?? - 1
Correct Answer: 9!/(2!)^{2}x3!
Explanation:
n items of which p are alike of one kind, q alike of the other, r alike of another kind and the remaining are distinct can be arranged in a row in n!/p!q!r! ways.
The letter pattern 'MESMERISE' consists of 10 letters of which there are 2M's, 3E's, 2S's and 1I and 1R.
Number of arrangements =
Correct Answer: 600
Explanation:
Two men, three women and one child can be selected in ?C? x ?C? x ?C? ways = 600 ways.
Correct Answer: 5040
Explanation:
Given word is GLACIOUS has 8 letters.
=> C is fixed in one of the 8 places
Then, the remaining 7 letters can be arranged in 7! ways = 5040.
Correct Answer: 60
Explanation:
Here given the required digit number is 4 digit.
It must be divisible by 5. Hence, the unit's digit in the required 4 digit number must be 0 or 5. But here only 5 is available.
x x x 5
The remaining places can be filled by remaining digits as 5 x 4 x 3 ways.
Hence, number 4-digit numbers can be formed are 5 x 4 x 3 = 20 x 3 = 60.
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