InfyTQ Permutation and Combination Quiz 3

In the given word LEADING, there are 3 vowels - E, A & I & we want them together.

As this is a word with 7 letters, we can choose consecutive places in 5 ways.

3 vowels can change their places in 3! ways.

Other 4 consonants can reverse themselves in 4! ways.

Total ways = 5 × 3! × 4! = 720.

If a man is Red, his mother must be red, his mothers brother also red but after
marriage, he gets converted to Brown.

consider 4 digits _ _ _ _
1st blank can be filled in ^{7}\textrm{C}_{1}

ways (0,3,6 are neglected as the first digit should
not be 0)
2st blank can be filled in ^{8}\textrm{C}_{1} ways (0 considered along with 1,2,4,5,7,8,9)
3st blank can be filled in ^{8}\textrm{C}_{1} ways
4st blank can be filled in

^{8}\textrm{C}_{1} ways
Therefore total 4 digit number without 3 and 6 is 7 x 8 x 8 x 8=3584

If only 12 people shaked their hands, then total hand shakes are

^{12}\textrm{C}_{2}

= 66

here, it is given that , there are two boxes , one containing 39 red balls and other one containing 26 green balls.
Also, we are allowed to move the balls.
therefore, we keep only one red balls in first box and move all the remaining balls in 2nd box.
So, 1st box contains only 1 red ball , 2nd box contains 26 green+38 red 64 balls.
Probability of choosing any box is \frac{1}{2}​ = 0.5
So, probability of taking one red ball is = \frac{1}{2}\times 1 +\frac{1}{2}\times \frac{38}{64}

= 0.5 + 0.29 = 0.79

: First card can go into any of the five boxes, Second can go into any of the
five
boxes, Third can go into any of the five boxes = 5×5×5=125

We take two odd numbers as (2n + 1) and (2n - 1).
Their sum should be less than 100. So (2n + 1) + (2n - 1) < 100 ⇒ 4n < 100.
The largest 4 multiple which is less than 100 is 96

Total ways of selecting two boxes out of 4 is^{4}\textrm{C}_{2}

= 6. Now, the number of ways of
selecting
two boxes where none of the green or red box included is only 1 way. (we select
yellow and
blue in only one way). If we substract this number from total ways we get 5 ways.

3450=2^{3}\times 3^{4}\times 5^{1}=a×b×c
We have to distribute three 2's
to a, b, c in^{3+3-1}\textrm{C}_{3-1} =^{5}\textrm{C}_{2} =10 ways
We have to distribute four 3's to a, b, c
in ^{3+4-1}\textrm{C}_{3-1}

=^{6}\textrm{C}_{2}

=15 ways
We have to distribute one 5 to a, b, c
in 3 ways.
Total ways = 10×15×3 = 450 ways.