Question

Number of words which can be formed using all the letter of the word HIP HIP HURRAY in which all H's lie somewhere between R's is

Answer 1

@rational

Answer 2

There are `12` letters in total
```
H : 3
I : 2
P : 2
U : 1
R : 2
A : 1
Y : 1
```

Answer 3

Put this string of 5 letters in a bag : `RHHHR`

Answer 4

then we have \(12-5+1 = 9\) objects to arrange, out of which I and P repeat 2 times :
\[\frac{9!}{2!2!}\]

Answer 5

read the question again

Answer 6

Oops! the H's need not be adjacent okay

Answer 7

yep..i thought of making cases but that would be silly... and I did a question in the past and i do have a bare hint of that
arranging some letters and rest the letters will be arranged by themselves or something like that

Answer 8

leave that question...
got a new one
If a number is selected at random from the set of all five digit numbers in which sum of the digits is equal to 43, then find the probability that this number is divisible by 11 is

Answer 9

@perl

Answer 10

\[
a=\sum_{k=0}^4a_k10^k\equiv0\pmod{11}\iff\sum_{k=0}^4(-1)^ka_k\equiv0\pmod{11}
\]
{9,9,9,9,7}, {9,9,9,8,8} are the only possible combinations of digits such that they sum to 43.
\[9-7+9-9+9=11\equiv0\pmod{11}\]
And my iPad is running out of battery so I will leave the rest for you

Answer 11

um quite confused!

Answer 12

a number is divisible by 11 when alternate digits sum is equal and idk how to arrange 99988

Answer 13

For {9,9,9,8,8}, there are only three situations. Either two 8's are negative, one 8 is positive and one 8 is negative or two 8's are positve. So sum of digits are either 3*9-2*8=11, 9-9+9-8+8=9 or 8-9+8-9+9=7.

For {9,9,9,9,7}, 9-7+9-9+9=11 is only solution.

Answer 14

So 98989 is the only number that is divisible by 11 formed by the set of digit {9,9,9,8,8} as 9-8+9-8+9=11. The only valid exchange of digits such that the number will still be divisible by 11 is switching the two 8's, which will result in the same number.

For {9,9,9,9,7}, 97999 and 99979 are both valid solution as 9-7+9-9+9=9-9+9-7+9=11.

So there are three solutions. You should be able to figure out the rest.

Answer 15

one more doubt... :D
how to examine if a number is divisible by 11?
what i have in mind is alternate number sum should be equal like 121; 1+1 =2
so I dont get what do you mean by two 8's are negative

Answer 16

By two 8's are negative, I mean that in 9-8+9-8+9 the two 8's are subtracted from other numbers.

Answer 17

@thomas5267 got any idea for the original permutation question?

Answer 18

Do the words formed have to be valid in English or is this just a combinatorial question?

Answer 19

H : 3
R : 2
I : 2
P : 2
U : 1
A : 1
Y : 1

Consider the case where R are at both ends of the word. Then there are \(\dfrac{10!}{3!2!2!1!1!1!}=151200\) possibilities.

I postulate that the combinations are equal to\[
\frac{7!}{2!\,2!}\sum_{k=0}^7\binom{n+3}{3}(8-n)=997920
\]

Answer 20

Let me explain the divisible by 11 question.

Part a, the set of digits such that the sum is 43.

The maximum sum of digit of a 5 digit number is 45.
\[
9+9+9+9+9=45\\
9+9+9+9+9-2=43\\
9+9+9+9+7=43\\
9+9+9+(9-1)+(9-1)=43\\
9+9+9+8+8=43
\]
So the two sets are {9,9,9,8,8} and {9,9,9,9,7}.

Part b, numbers that are divisible for 11 constructed using the two sets of digits.
\[
a=\sum_{k=0}^4a_k10^k\\
a=\sum_{k=0}^4a_k10^k\equiv0\pmod{11}\iff\sum_{k=0}^4a_k(-1)^k\equiv0\pmod{11}\\
a_0-a_1+a_2-a_3+a_4\equiv0\pmod{11}
\]
Consider the set {9,9,9,8,8}. We have to add three elements of the set and subtract two elements of the set such that the result is 11n (as \(a_0-a_1+a_2-a_3+a_4\equiv0\pmod{11}\)). Either we add two 8's, we add one 8 and subtract one 8, or we subtract two 8's.

8+8-9+9-9=7 (add two 8's)
9-9+9-8+8=9 (add one 8 and subtract one 8)
9+9+9-8-8=11 (subtract two 8's)

So 98989 is the only number constructed using {9,9,9,8,8} as digits that is divisible by 11. (We have subtracted two 8's, so it must be in \(a_1\) and \(a_3\). Swapping the two 8's still yields 98989.)

Consider the set {9,9,9,9,7}. Again, add three elements and subtract two such that the results equal 11n. This time, we either add 7 or subtract 7.

9+9+7-9-9=7 (add 7)
9+9+9-9-7=11 (subtract 7)

So 7 is either \(a_1\) or \(a_3\) as we have subtracted 7. 99979 (\(7=a_1\)) and 97999 (\(7=a_3\)) are the two numbers.

There are 3 numbers that are divisible by 11 in the set of five digits numbers such that the sum of digits = 43. The probability of finding such number in the set is left as an exercise for the reader XD.