Question

If number of arrangements of letters of the word "DHARAMSHALA" taken all at a time so that no two alike letters appear together is $(4^a.5^b.6^c.7^d)$, (where $a,b,c,d \in N$), then $a+b+c+d$ is equal to

Answer 1

7

Answer 2

The word "DHARAMSHALA" has 10 letters: D(1), H(1), A(3), R(1), M(1), S(1), L(2). We need to find the number of arrangements where no two identical letters are together. This requires the Principle of Inclusion-Exclusion.

Total arrangements of "DHARAMSHALA": Total letters = 10. 'A' repeats 3 times, 'L' repeats 2 times. Total arrangements = $\frac{10!}{3!2!} = \frac{3628800}{6 \times 2} = 302400$.

Let $P_{AA}$ be the property that at least two 'A's are together (AA). Let $P_{LL}$ be the property that at least two 'L's are together (LL).

We want to find the number of arrangements with neither $P_{AA}$ nor $P_{LL}$. Using PIE: Total - $|P_{AA}| - |P_{LL}| + |P_{AA} \cap P_{LL}|$.

1. $|P_{AA}|$: Treat "AA" as a single unit. We arrange {(AA), A, D, H, R, M, S, L, L}. This is 9 items with 'L' repeated twice. $|P_{AA}| = \frac{9!}{2!} = \frac{362880}{2} = 181440$.

2. $|P_{LL}|$: Treat "LL" as a single unit. We arrange {D, H, A, A, A, R, M, S, (LL)}. This is 9 items with 'A' repeated thrice. $|P_{LL}| = \frac{9!}{3!} = \frac{362880}{6} = 60480$.

3. $|P_{AA} \cap P_{LL}|$: Treat "AA" as one unit and "LL" as another. We arrange {(AA), A, D, H, R, M, S, (LL)}. This is 8 distinct items. $|P_{AA} \cap P_{LL}| = 8! = 40320$.

Number of arrangements with no two alike letters together = Total - $|P_{AA}| - |P_{LL}| + |P_{AA} \cap P_{LL}|$ $= 302400 - 181440 - 60480 + 40320$ $= 302400 - 241920 + 40320 = 60480 + 40320 = 100800$.

We are given this number as $(4^a.5^b.6^c.7^d)$. Prime factorization of $100800$: $100800 = 1008 \times 100 = (16 \times 63) \times 100 = (2^4 \times 3^2 \times 7) \times (2^2 \times 5^2) = 2^6 \times 3^2 \times 5^2 \times 7^1$.

The given form in prime factors: $4^a.5^b.6^c.7^d = (2^2)^a \cdot 5^b \cdot (2 \cdot 3)^c \cdot 7^d = 2^{2a} \cdot 5^b \cdot 2^c \cdot 3^c \cdot 7^d = 2^{2a+c} \cdot 3^c \cdot 5^b \cdot 7^d$.

Comparing exponents: For base 3: $c = 2$. For base 7: $d = 1$. For base 5: $b = 2$. For base 2: $2a+c = 6$. Substituting $c=2$: $2a+2 = 6 \implies 2a = 4 \implies a = 2$.

So, $a=2, b=2, c=2, d=1$. These are all natural numbers. The required sum is $a+b+c+d = 2+2+2+1 = 7$.