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 =
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
It 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 balls
Hence, 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
=
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
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.
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
Two men, three women and one child can be selected in ?C? x ?C? x ?C? ways = 600 ways.
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.
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.
n(S) = 52C3 = 132600/6 = 22100
n(E) = 4C3 = 24/6 = 4
A person can be chosen out of 18 people in 18 ways to be seated between Musharraf and Manmohan. Now consider Musharraf, Manmohan, and the third person, sitting between them, as a single personality, we can arrange them in 17! ways but Musharraf and Manmohan can also be arranged in 2 ways.
Required number of permutations = 18 x (17!) x 2 = 2 x 18!
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