1) How many 9 letter palindromes are possible from English Alphabets?
2) How many integers are there between 1 and 10^4 that contain exactly one 8 and one 9?
Cheers
1)i just have to choose letters for the first 5 places coz for the word to be a palindrome once the first four letters are chosen the last four are also automatically fixed. now the first letter can be chosen in 26 ways,the second in 26 ways and so on upto the 5th letter(allowing repetitions). so total no.of ways allowing repetitions is 26^5 without repetions the first letter can be chosen in 26 ways,2nd letter in 25 ways,3rd in 24,4th in 23,5th in 22 ways. without repetitions the total no.of ways is 26*25*24*23*22=P(26,5)
2)since we have to have both 8 and 9 we will first fix their positions.this can be done in P(4,2) since we consider all numbers from 1(0001) to 9999(we need not consider 10000).the remaining 2 digits must be filled with digits from 0 to 7 since we can have only one 8 and one 9.this can be done in 8*8 ways. so total no. of ways is 8*8*P(4,2).
In the palindrome ABCDEDCBA, it is not specified that A,B,C,D,E are necessarily different it could be like AAAAAAAAA where A is any alphabet Hence 26^5.
In the palindrome ABCDEDCBA, it is not specified that A,B,C,D,E are necessarily different it could be like AAAAAAAAA where A is any alphabet Hence 26^5.
I think capricious is very correct on this... What do u think pendyal : :
There are 7 distinct coins which r 2 b distributed randomly into 5 distinct pots in a row. Any pot can receive any no. of coins. All coins need to b distributed. In how many ways can these coins b distributed so that no pot is empty?
There are 7 distinct coins which r 2 b distributed randomly into 5 distinct pots in a row. Any pot can receive any no. of coins. All coins need to b distributed. In how many ways can these coins b distributed so that no pot is empty?
There are 7 distinct coins which r 2 b distributed randomly into 5 distinct pots in a row. Any pot can receive any no. of coins. All coins need to b distributed. In how many ways can these coins b distributed so that no pot is empty?
Can some1 explain using the grouping approach
The No. ways in which first 5 coins can be selected out of 7 = 7C5 The No. ways in which the selected five coins can be put in five different pots = 5x4x3x2x1 The No. ways in which the remaining two coins can be placed in 2 of five pots = 5x5 Thus Ans = 7C5 x (5x4x3x2x1) x 5x5 = 7P2 x (5x5)
Q - Ben and Ann are among 7 contestants from which 4 semifinalists are to be selected. Of the different possible selections, how many contain neither Ben nor Ann?
Q - Ben and Ann are among 7 contestants from which 4 semifinalists are to be selected. Of the different possible selections, how many contain neither Ben nor Ann?
If the answer is 5 the problem is quite simple.
You need to select the 4 semifinalists from the remaining 5 contestants.
Which can be done in 5C4 ways = 5 ways
But if it means that both are not simultaneously selected then the answer wud be
7C4 - 5C2 = 35 - 10 = 25 ways
Here 7C4 are total no. of ways 4 semifinalists are selected out of 7 and 5C2 are ways in which both Ben and Ann are selected.
Hi all, Could any one of you please give me the solution for the question below.This is eating my brain. There are 5 letters and 5 distinct addressed envelopes. In how many ways can these letters go into wrongly addressed envelopes. Kindly post the procedure also.
Hi all, Could any one of you please give me the solution for the question below.This is eating my brain. There are 5 letters and 5 distinct addressed envelopes. In how many ways can these letters go into wrongly addressed envelopes. Kindly post the procedure also.
1 Right can go in 45 ways. 2 Rights = 20 3 Rights = 10 4 Rights = 0 5 Rights = 1 0 Rights = Have to Find out !!
Total ways shud be 5! = 120
==>> o Rights ( or all Wrongs ) = 120 - ( 45 + 20 + 10 + 0 + 1 ) = 120 - 76
This is my first post in this thread i am having some confusion on one of the questions which i am mentioning below... Please mention the different approaches that can be taken to arrive to the solution of this problem... Q:In how many ways can 10 identical presents be distributed among six children so that each child gets atleast 1 present?
This is my first post in this thread i am having some confusion on one of the questions which i am mentioning below... Please mention the different approaches that can be taken to arrive to the solution of this problem... Q:In how many ways can 10 identical presents be distributed among six children so that each child gets atleast 1 present?
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