the unlimited potential that lies within you. let it slumber no more.
There are 6 boxes numbered 1, 2 ...6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is
We have to use at least 1 green ball.
So we can use 1, 2, 3, 4, 5 or 6 green ball.
If we use 1 green ball then the probable way is 6C1=6
If we use tow green ball then as the two box will have to be consecutively numbered for this reason we have to take this two ball as a single number and the probable way will be 5C1=5
and same way if we use 3 green ball then 4C1=4 and so on
Ten coins are tossed simultaneously. In how many of the outcomes will the third coin turn up a head?
Just ten coin tossed simultaneously and we have to find the number of way in we can get third coin turn up as head. The rest 9 coin can be turn up either head or tail. If we toss n coin simultaneously then the probable outcome is 2n. So for 9 coin it is 29.
Example for two coin, HH, HT, TH, TT= 4 = 22
For three coin, HHH,HHT,HTH,THH, HTT,THT,TTH,TTT=8=23
How many different positive integers exist between 10^6and 10^7, the sum of whose digits is equal to 2?
Existing positive integer will be in between 106 =1000000 and 107 =10000000.
As the sum of the digit is two so the number may start only with 1 or 2.
If it start with 1 then there will be one more 1 and the rest will be zero.
As it start with 1 and there are more 6 position the rest 1 may sit in 6C 1=6 way.
If it start with 2 then the rest digit will be zero. So here we will get only one number
Here 4! is the total way of arrangement taking the vowel as a single letter.
(3!/2!) is the way of arrangement of the three vowel
If in a word contain n letter and if among them, there is some letter repeat p, q, ........r times then total way of combination is
Here total way of arrangement will be 11!/(2!*2!*2!) as A, M and T repeat 2 times
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